Kinematics (Integreated) (Aumr Garu) (124-177)

KINEMATICS PHYSICS - I A
KINEMATICS
5
Albert Einstein (14 March 1879 – 18 April 1955) was a German-born theo-
retical physicist who discovered the theory of general relativity, effecting a revolu-
tion in physics. For this achievement, Einstein is often regarded as the father of
4
CHAPTER
5
modern physics. He received the 1921 Nobel Prize in Physics "for his services to
theoretical physics, and especially for his discovery of the law of the photoelectric
effect".
He continued to deal with problems of statistical mechanics and quantum
theory, which led to his explanations of particle theory and the motion of molecules.
He also investigated the thermal properties of light which laid the foundation of the
photon theory of light. In 1917, Einstein applied the general theory of relativity to
model the structure of the universe as a whole. Albert Einstein
h Straight line motion h Oblique projectile

h Motion under gravity h Horizontal projectile
4.1 INTRODUCTION (numerical value) and no direction. whereas, the

The study of motion of bodies without displacement has magnitude as well as direction.
considering the cause of motion is called Hence, distance is scalar and displacement is vector.
Kinematics. In the universe every object moves, Application 4.1: Consider a particle moving
though some objects appear to be stationary. The from A to B along a curve as shown.
terms motion and rest are relative. We begin this
chapter with motion along a straight line, i.e. motion
in one dimension. Later, we look into motion in a
A B
plane, i.e. two-dimensional motion.
The distance travelled is equal to the length of
4.2 REST, MOTION the curve AB, whereas the magnitude of the
If a particle’s position does not change either displacement is equal to the length of the straight
with respect to a fixed point and with respect to time, line AB. C
it is said to be at rest.
5m
If a particle’s position is continuously changing 4m
with respect to a fixed point and with respect to time, Application :4.2
it is said to be in motion. A B
3m
4.3 MOTION IN A STRAIGHT LINE If a person walks from A to B and then from B
When a particle is in motion, if the line joining to C as shown,
the successive positions is a straight line then it is Distance travelled = AB + BC= 7m
said to be in straight line motion. Here, the reference Displacement = AC = 5m
point about which the position of the particle is
located is called origin. Application 4.3 : A particle moves over an
4.4 DISTANCE AND DISPLACEMENT arc PQ , of a circle of radius R, subtending an angle
“Length of actual path between initial and  at the centre .
P Q
final positions is called distance”.  R
R
“The shortest straight line distance directed O
from initial position to final position irrespective
of the path is called displacement”.
a) distance travelled = arc PQ = R
The ‘distance’ and ‘displacement’ are two
 
different quantities. Distance has just a magnitude b) displacement= straight line PQ = 2R sin  
 2
AKASH MULTIMEDIA 124
PHYSICS - I A KINEMATICS
Note 4.1 : If a particle starts from a point and 4.8 NON – UNIFORM MOTION OR
reaches the same point at the end of its journey, then VARIABLE MOTION
displacment is zero. However distance covered is If a particle moving along a straight line travels
not zero. Therefore, a particle can travel some unequal distances in equal intervals of time or equal
distance without displacement.
distances in unequal intervals of time, we say that
Note 4.2 : For a particle in motion,the magnitude the particle is in non-uniform motion.
of displacement cannot exceed the distance Ex : i) Motion of a freely falling body.
displacem en t < d istance (for curved motion).
1 2
displacem en t = distance (for staright line The equation of the motion is y  gt .
2
motion). The position–time graph is a parabola with
4.5 SPEED increasing slope.
The distance travelled by a body in unit time
is called it’s speed.
y
It is a scalar quantity.
CGS unit is cms–1 and SI unit is ms–1.
4.6 VELOCITY o t
The displacement of a body in unit time is Position time graph of a freely falling body in non
called it’s Velocity. - uniform motion.
It is a vector quantity. Ex : ii) Motion of a body thrown vertically
CGS unit is cms–1 SI unit is ms–1. upwards.
4.7 UNIFORM MOTION 1
The equation of motion is y  ut  gt 2
If a particle moving along a straight line (say x– 2
The position–time graph is a parabola.
axis) travels equal distances in equal intervals of time
Its slope decreases from + K to zero and
we say the particle is in uniform motion. The motion
thereafter from zero to – K, where K is instantaneous
is expressed by an equation of the form.
slope.
x = vt + b
where x is the position coordinate of the particle,
t is the time, v and b are certain constants. In this y
equation position is a linear function of time. Hence

the slope of the position - time graph is a straight
o t
line, the slope of which is a constant( v) and is equal
Position time graph of a body thrown
to constant velocity or uniform velocity of particle.
vertically upwards.
x
4.9 UNIFORM SPEED
If a body travels equal distances in equal
b
intervals of time however small the intervals may
o t be, then it is said to be moving with uniform speed.
Position time graph of an object in uniform
4.10 NON – UNIFORM SPEED
motion. If a body travles equal distances in unequal
When t = 0, the above equation can be written as intervals of time or unequal distances in equal intervals
x0 = b
Where x0 indicates the initial position of the of time then it is said to be moving with non-uniform
particle from the origin. speed.

4.11 INSTANTANEOUS SPEED from B to A in the same path with speed v2 . The
The speed of a particle at a particular instant of 2v1v 2
average speed of total motion is
time is called it’s instantaneous speed. v1 + v 2
If S is the distance travelled by a particle in a Application 4.6 :
A body is travelling between two positions. The
time interval t then
total distance is divided into n equal parts. These parts
S are travelled with speed v1, v2, v3.....vn respectively.
speed =
t The average speed of total motion is such that
If the time interval t is chosen to be very small, n 1 1 1 1
   ......... 
i.e., as t  0 , then the corresponding speed is called Average speed v1 v2 v3 vn
instantaneous speed. Application 4.7:
S dS A body travelling between two positions traveles
 Lt   instantaneous speed. with speed v1 for time t1 and then with speed v2 for
t  0 t dt
time t2. For the total motion,
4.12 AVERAGE SPEED v t +v t
Average Speed = 1 1 2 2
For a particle in motion (uniform or non- t1 + t 2
uniform), the ratio of total distance travelled to the Application 4.8:
total time of motion is called average speed. A body travelling between two positions travels
Total distance first half of the time with speed v1 and the next half
Average speed = of the time with speed v2. The average speed of total
Total time
If s1, s2, s3......sn are the distances travelled by a v1 + v 2
motion is
particle in the time intervals t1, t2, t3......tn respectively 2
then, Application 4.9 :
s  s  s ......sn A body travelling between two positions travels
Average Speed  1 2 3 for the time intervals t1, t2, t3..........tn with speeds v1,
t1  t 2  t 3 .......t n
v2, v3,...........vn respectively
Application 4.4 :
A body travelling between two positions traveles total distance
Avg. speed =
first half of the distance with speed v1 and the next total time
half of the distance with speed v2 The average speed
2v1v2 v1t1 + v2 t 2 + v3 t 3 +............v n t n
of total motion is v + v =
1 2
t1 + t 2 + t 3 +.........t n
Let x be the total distance between two positions. 4.13 UNIFORM VELOCITY
Let t1 be the time for first half and t2 be the time If a body has equal displacements in equal
for the next half of the distance intervals of time however small the intervals may be
Total distance then it is said to be moving with uniform velocity.
Avg. speed = 
Total time 4.14 NON-UNIFORM VELOCITY
If a body has equal displacements in unequal
x x 2 v1v 2
  intervals of time or unequal displacements in equal
t1  t 2 x x v1 + v 2 intervals of time then it is said to be moving with

2 v1 2 v2 Non–Uniform Velocity.
Application 4.5: Note : The displacement variation may be due
A body is travelling between two positions A, to change in magnitude or change in direction of
B. It travelles from A to B with speed v1 and then motion or both.

4.15 INSTANTANEOUS VELOCITY Note 4.8 : In uniform motion, the instantaneous

The velocity of a particle at a particular instant velocity of a body is equal to the average velocity.
of time is called it’s instantaneous velocity. 4.17 ACCELERATION
If  S is the displacement by a particle in a time If the velocity of a particle is changing as it
interval t then  moves then it is said to be moving with acceleration.
 S
Velocity  V  The acceleration measures how rapidly the velocity
t is changing.
If the time interval t is chosen to be very small,
Acceleration is defined as the rate of change
i.e., as t  0 , the corresponding velocity is called of velocity.
instantaneous velocity.  
Let V1 , V2 be the velocities of a particle at
 
S dS instances t1, t2 respectively. Now,
Lt   instantaneous velocity
t  0 t dt change in velocity
The instantaneous velocity is rate of change of Acceleration 
position with time.  time 
 V2 - V1 DV
a = 
ds s2 t2
t 2 - t1 Dt
Note 4.3 : v  ;  ds   vdt
dt s1 t1 It is a vector. It is in the direction of change
4.16 Average velocity : in velocity.
For a particle in motion (uniform or non- S.I. Unit is ms–2, dimensional formula is [L1T–2]
uniform), the ratio of total displacement to the total Note 4.9 : The velocity variation may be due to
time interval is called Average velocity. change in magnitude of velocity (speed) or change
Total displacement in direction of velocity. Hence acceleration may be
Average velocity =
Total time due to either of the above reasons or both.
Suppose a particle  displaces from P1 to P2 in a eg 1 : For a car going on a straight road if the
time
 interval  t. If x1 is initial position vector and speed is increasing, then the acceleration is due to
x 2 is final position vector then change in magnitude of velocity
 
 x 2  x1 eg 2 : For a stone whirled in a horizontal circle
Average velocity = v 
t with constant speed, the acceleration is due to change
Note 4.4 : A particle travelling between two in direction of velocity.
positions A, B travels from A to B with velocity v1
eg 3 : For a stone whirled in a vertical circle
and returns from B to A with velocity v2. Now
with a changing speed, the acceleration is due to
average velocity of total motion is zero since the net change in both magnitude and direction of velocity.
displacement of the particle is zero.
Note 4.10 : The acceleration of a moving particle
Note 4.5 : In variable motion, the average
may be positive or negative. If the speed of particle
velocity depends on the interval of the time during
is increasing with time then acceleration is positive
which the velocity is calculated.
and if the speed is decreasing with time then
Note 4.6 : In uniform motion, the average
acceleration is negative.
velocity is a constant and is same for all the intervals
of time during which the value is calculated. Note 4.11 : For positive acceleration the velocity
Note 4.7 : The magnitude of average velocity is vector and acceleration vector are in the same
equal to the average speed for motion along a straight direction. But for negative acceleration, the velocity
line and it is a scalar quantity. and acceleration vectors are opposite


4.18 DECELERATION OR RETARDATION final velocity after t seconds with x f as final position
If the speed is decreasing with time then vector.
acceleration is negative.
Now the equations of motion are as follows.
The negative acceleration is called decelera
tion or retardation. 1) Velocity as a function of time
  
4.19 INSTANTANEOUS ACCELERATION v  u  at (or)v = u + at
The acceleration of a particle at a particular 2) Displacement as a function of time
instant of time is called it’s instantaneous acceleration.     1  1
As the velocity of a particle is changing with S  x f  x i  ut  at 2 (or) S = ut + at 2
2 2
time, if the time interval approaches to zero (i.e., to 3) Position as a function of time
an infinitely short time interval)
 
   1  2 1
 V dV
x f  x i  ut  at (or) x f  x i  ut  at 2
Lt   Instantaneous acceleration 2 2
t  0  t dt
4) Velocity as a function of displacement
Note 4.12 : Instantaneous accelar ation    2 2
v.v  u.u  2 a.s (or) v –u = 2as
dv d 2s v2 t2
5) Displacement in n th second of motion
a  2 ;  dv   a dt
dt dt v1 t1
    1  1
dv ds dv Sn = u + a  n -  (or) Sn = u + a  n - 
Note 4.13 : a  . ; a  v.  2  2
ds dt ds
6) Displacement = (Average velocity)time
 ads   vdv
 
4.20 UNIFORM ACCELERATION   u  v  u  v
S t and S   t
If the average acceleration over any time interval  2 
  2 
equals the instantaneous acceleration at any instant
4.22 MOTION CURVES
of time then the acceleration is said to be uniform or
constant. It does not vary with time. The velocity Graphical analysis is a convenient method of
either increases or decreases at the same rate studying the motion of a particle. It can be effectively
throughout the motion. applied to analyse the motion situation of a particle.
The variations of two quantities with respect to one
4.21 KINEMATICAL EQUATIONS OF
another can be shown graphically.
MOTION OF A PARTICLE
MOVING ALONG A STRAIGHT LINE For graphical representation, we require two
WITH UNIFORM ACCELERATION : coordiate axes. The usual practice is to take the
Kinematical equations are useful to solve independent variable along X-axis and the dependent
problems in one dimensional motion of a particle variable along Y-axis. In a context, with time as one
with constant acceleration. of the variables, it is usually taken along x-axis (since
it is independent) and the other variable is along y-
Consider a particle with initial position vector
  axis.
x i . Suppose it starts with initial velocity u and

moves with uniform acceleration a. Suppose v is its

4.22(i) DISPLACEMENT – TIME GRAPHS (S - T GRAPHS)

Graphs are drawn with time along x-axis and displacement along y–axis
significance : i) The slope of the tangent at any point gives the instantaneous velocity.
ii) The slope of the chord between two positions gives average velocity.
Context Shape of graph Comment on shape of graph
y
S i) A straight line parallel to x-axis.

1) Particle at rest
ii) S0 indicates initial displacement
S0
X
O t
y
2) Particle with Uniform velocity,
. S = 0 at t = 0 S A straight line with positive slope
The equation of motion is
S = ut
X
O t
y
3) Particle with uniform
acceleration, S = 0 at t = 0
The equation of motion is S A parabola with increasing slope
1
S = ut + a t2
2 X
O t
y
4) Particle with uniform
acceleration, S = S0 at t = 0
The equation of motion is S i) A parabola with increasing slope
1 ii) Intercept on y – axis is initial displacement
S = S0 + ut + at2 S0
2 X
O t
y
5) Particle with uniform retardation
S = 0 at t = 0
The equation of motion is S A parabola with decreasing slope
1
S=ut + at2 where a is negative
2 X
O t
y
6) Particle projected vertically
Upwards. i) A parabola.
The equation of motion is S ii) Its slope decreases from +K to zero and
1 2 there after from zero to – K where K is
S = ut - gt instantaneous slope.
2 X
O t

4.22(ii) VELOCITY – TIME GRAPHS (V – T GRAPHS) :

Graphs are drawn with time along x-axis and velocity along y–axis
significance :
i) The slope of the tangent at any point gives the instantaneous acceleration.
ii) The slope of the chord between two positions gives average acceleration.
iii) The area of the figure bounded by the graph, the time axis and the ordinates drawn at the initial and
final positions on the time axis gives the displacement in the bounded time interval.

y
v As v = 0, the graph is a straight line

1) Particle at rest
along x-axis
X
O t
y
2) Particle with Uniform velocity (u) v i) A straight line parallel to x-axis
The equation of motion is u
ii) slope = 0 , acceleration = 0
V = constant = u iii) y intercept = initial velocity = u
X
O t
y
3) Particle with uniform acceleration,
v
with velocity = 0 at t = 0 i) A straight line with positive slope
The equation of motion is ii) Slope = acceleration = constant
V = at X
O t
y
4) Particle with uniform acceleration,
v i) A straight line with positive slope
with velocity = u at t = 0
ii) Intercept on y – axis = u
The equation of motion is u iii) Slope = acceleration = constant
V = u + at X
O t
y
5) Particle with uniform retardation, u
with velocity = u at t = 0 i) A straight line with negative slope
v
The equation of motion is ii) Slope = retardation = constant
V = u + at where a is negative X
O t
y i) A straight line with negative slope.

u ii) The net area bounded by the curve
6) Particle projected vertically is zero.
upwards v X iii) Net displacement in the time of
A B
The equation of motion is O flight = 0.
t
V = u – gt iv) OA = time of ascent = u/g.
AB = time of descent = u / g.
u OB = time of flight = 2u / g.

4.22(iii) ACCELERATION - TIME GRAPHS (A – T GRAPHS) :

Graphs are drawn with time along x-axis and acceleration along y–axis
Significance i) The slope of the tangent at any point gives instantaneous jerk.
ii) The area under the graph with x – axis (time axis) gives the change in velocity in the
bounded time interval (v – u)
y
a0 The graph is a straight line parallel to

1) The particle is with constant a x–axis
acceleration a0 Y intercept = acceleration = a 0
x
O t
y
2) Particle with non-uniform a

acceleration and The graph is with variable slope
with a = 0 at t = 0
x
t
4.23 GRAPHICAL TREATMENT 1
KINAMATICAL EQUATIONS OF 2) To Show that S = ut + at 2
2
MOTION OF A BODY WITH UNIFORM
The area under the velocity time graph with
ACCELERATION
X–axis (time axis) gives the displacement in the
Consider a particle moving with initial velocity
bounded time interval. Here the area bounded by
‘u’ and uniform acceleration ‘a’. Suppose ‘v’ is its
the line AB with x-axis gives the displacement
velocity after ‘t’ seconds. Let ‘S’ be its displacement
in the time interval t. The velocity time graph is a \ S = Area of rectangle OACD + Area of triangle
straight line with positive slope. The graph is given ABC
by the line AB. 1
\ S = (OA)(OD)+ (AC)(CB)
v
y 2
B 1
 = (u )(t )+ (t )(v - u )
Velocity v-u
2
 1 æ v - u ö÷
A
t C = ut + (t )(at ) ççç a = ÷
(u) è t ø÷
2
1 2
x \ S = ut + at
O D 2
time
Note 4.14 :
1) To show that v = u + at 1
General Method to show that S  ut  at 2
The slope of velocity time graph gives the 2
acceleration of the particle. When the particle is moving with uniform
acceleration,
BC v - u
Here, slope = tan  = = =a u+v
AC t Average Velocity =
2
 v - u = at (or) v = u + at
Displacement = (Average Velocity) (time) SOLVED PROBLEMS BASED ON

u+v KINEMATICAL EQUATIONS
S 
 2  t 
  Problem : 4.1
 u +u + at  The displacement x of a particle at the instant when
S 
 2
  t   V = u + at 
  its velocity v is given by v= 3 x + 16 . Find itss
1 2 acceleration and intial velocity
 S  ut + at
2
Sol. v = 3 x + 16 or v2 = 3x + 16 or v2 - 16 = 3x Comparing
2 2
Note 4.15 : To show that v - u = 2aS with v2 - u2 = 2aS , we get , u = 4 units,2a= 3 or a = 1.5
v = u + at  v  u  at ----- (1)
units
u+v 2S
S= t v + u = ----- (2) Problem : 4.2
 2  t If Sn = 2 + 0.4n find intial velocity and acceleration
Using (1), (2)
 2S  Sol. Sn = 2 + 0.4n
(v – u) (v + u) = (at)   = 2aS 0.4 0.4
 t  Þ Sn = 2 + 0.4n - +
2
 v - u = 2aS 2 2 2
Note 4.16: æ 1ö
Þ Sn = 2.2 + 0.4 ççn - ÷ ÷
çè 2 ø÷
Motion of a body
æ 1ö
Compairing it with Sn = u + a çççèn - ø÷
÷
constant 2÷
velocity Accelerated U = 2.2 units a = 0.4 units
motion
s = ut
Problem : 4.3
The two ends of a train moving with uniform
uniformly non-uniformly
accelerated accelerated acceleration pass a certain point with velocity u and
v. Find the velocity with which the middle point of
ds dv dv the train passes the same point.
v = u + at v= ;a = = v u
dt dt ds v1 v
Sol.
differentiation s/2 s/2
æu + v ö÷
s = çç ÷t s v a v - u 2 = 2a s / 2
12
çè 2 ø÷ Integration v 2 - v12 = 2as / 2
v2–u2 = 2as ò ds = ò vdt Þ v12 - u 2 = v 2 - v12
2v12 = v 2 + v 2
1
S = ut + at 2 ò dv = ò a dt u 2 + v2
2 v1 =
2
æ 1ö
Sn = u + a ççn - ÷ ÷ ò ads = ò vdv Problem : 4.4
çè 2 ø÷ The velocity of a body moving with uniform
Note 4.17: For two bodies in motion for same time acceleration of 0.3m/s2 is changed to 3m/s in certain
interval we can use equation of motion in the time. If the average velocity in the same time is 30m/
relative form such as s then find the distance travelled by it in that time.
Sol. v –u = at
vr = ur + ar t 3 = 0.3t Þ t = 10s
1 S = Vaverage x time
S r = ur t + ar t 2 and vr2 - ur2 = 2ar sr
2 = 30 ´ 10 = 300 m

Problem : 4.5 Problem : 4.8

A body starts from rest and moves with uniform A particle is at x = + 5 m at t = 0, x = – 7 m at t = 6 s
acceleration of 5 ms-2 for 8 seconds. From that time and x = + 2 m at t = 10s. Find the average velocity of
the acceleration ceases. Find the distance covered in the particle during the intervals
12s starting from rest. (a) t = 0 to t = 6s (b) t = 6s to t = 10s,
Sol. The velocity after 8 seconds v = 0 + 5× 8 = 40 m/s (c) t = 0 to t = 10 s.
Distance covered in 8 seconds From the definition of average velocity
1 x x2  x1
s0 = 0 + ´ 5´ 64 = 160 m v 
2 t t 2  t1
After 8s the body moves with uniform velocity and
(a) The average velocity between the times t = 0 to t = 6s
distance covered in 4s with uniform velocity
v=vt= 40 × 4 = 160 m
x1  5m, t1  0, x2  7 m t2  6 s
The distance covered in 12 s =160 +160 = 320 m. x2  x1 7  5
Hence v1    2ms 1
t2  t1 60
Problem : 4.6
(b) The average velocity between the times
A scooter can produce a maximum acceleration of t2 = 6s to t3 = 10 s is
5m s-2 . Its brakes can produce a maximum retardation
of 10 ms-2 . The minimum time in which it can cover a x3  x2 2   7  9
v2     2.25ms 1
distance of 1.5 km is ? t3  t 2 10  6 4
Sol. If v is the maximum velocity attained, then (c) The average velocity between times t1= 0 to t3=10 s is
v2 - O2 = 2´ 5´ S1 . Also , O2 - v2 = 2´ 10´ S 2 x3  x1 2  5
v3    0.3ms 1
v2 v2 t3  t1 10  0
S1 = , S2 =
10 20 Problem : 4.9
2 2 2
v v 3v Velocity and acceleration of a particle at time t = 0 are
S = S1 + S2 Þ 1500 = + = or  
10 20 20    
u  2iˆ  3 ˆj m / s and a  4iˆ  2 ˆj m / s 2 respectively.
1500´ 20 Find the velocity and displacement of particle at
v2 = = 10000 or v = 100 m s-1
3 t = 2s.
Problem : 4.7 Sol. Here, acceleration

The speed of a train is reduced from 60 km/h, to  
a  4iˆ  2jˆ m/s2 is constant. So, we can apply
15 km/h, whilst it travels a distance of 450 m. If the      1 2
retrardation is uniform , find how much further it will v  u  at and s  ut  at
2
travel before coming to rest ? substituting the proper values, we get
5 50 
Sol. Here u = 60´
18
=
3
m/s     
v  2iˆ  3jˆ  (2) 4Iˆ  2Jˆ  10Iˆ  7Jˆ m/s 
 1 2 ˆ
v = 15´
5
=
25
m/s 
and s   2 2iˆ  3ˆj   2
   
 2 4I  2Jˆ  12Iˆ  10Jˆ m
18 6
Therefore, velocity and displacement of particle at t =
Using v2 = u2 + 2as , we get
2s are 10iˆ  7jˆ  m/s and 12iˆ  10ˆj  m respectively..
2 6
æ50 ö÷ æ25 ö÷
çç ÷ = çç ÷ + 2´ 9´ 450 Problem : 4.10
çè 3 ø÷ çè 6 ø÷
A rifle bullet loses 1/20th of its velocity in passing
125 through a plank. What will be the least number of
or a = - m / s2
36 ´ 12 such planks required to just stop the bullet ?
2
If s1 is the further distance travelled before coming to æ19v ö÷
Sol. çç ÷ - v 2 = 2ax 02 - v2 = 2anx
rest, then çè 20 ø÷
- v2 1
v2 25 25´ 36´ 12 Dividing , n = =
s1 = = ´ = 30m æ19 ö÷
2
æ ö
çç v÷ - v 2 1- çç19 ÷
2
2a 6 6´ 2´ 125 çè 20 ø÷ ÷
çè 20 ø÷

20´ 20 400 Problem : 4.13

= = = 10.3
(20 + 19)(20 - 19) 39 A body starts with initial velocity u and moves with
uniform acceleration.when the velocity has
 11 planks so, the bullet shall stop in 11 th plank. increased to 5u, the acceleration is reversed in
Problem : 4.11 direction, the magnitude remaining constant. Find
its velocity when it returns to the starting point?
A car starting from rest, accelerates at the rate of f
through a distance S, then continues at constant speed u a a
A B
for time t and declerates at the rate v = ft1 to come to Sol. AB = S B C
A|
rest. If the total distance travelled is 15 S then S = V 5u
f v v f/2 For AB 25 u2 – u2 = 2as
Sol. A B t C D V 2 - 25u 2 = 2 (- a )´ (- s)
t1 2t 2 Þ V 2 - 25u 2 = 24u 2
15 S
V2 = 49u2 V= ± 7u
1 2
AB= S = f t1 -------- (1) v = ft1 \ V is opposite to u V = - 7u
2
BC = (ft1 )t Problem : 4.14
2 A train starting from rest travels the first part of its
u2 (ft1 ) journey with constant acceleration a, second part with
CD = =
2a 2 (f / 2) constant velocity v and third part with constant
retardation a , being brought to rest. The average
Sn = 2 + 0.4m
7v
S + f t1 t +25 =15 S speed for the whole journely is . For what fraction
8
ft1t = 12 S -------- ( 2 ) of the total time, the train travels with constant
velocity ?
Dividing (1) by (2)
1 1
t vt + vt 1 + vt
t1 = Sol. 7v 2 2 7 t + t1
6 = or =
2
8 t + t1 + t 8 2t + t 1
1 t2 ft
S= f \ S= or 6t = t1
2 36 72
t 6t 3
Problem : 4.12 Now = =
2t + t 1 8t 4
A body covers 100cm in first 2seconds and 128cm in Problem : 4.15
the next two seconds moving with constant
acceleration. Find the velocity of the body at the A particle traversed half of the distance with a
end of 8sec? veloicty of V0. The remaining parts of the distance
was covered with velocity V, for half of the time and
1 with V2 for other half of the time. Find the mean
Sol. 100 = 2u + a.4 -------(1)
2 velocity of the particle averaged and the whole time
1 of motion
228 = 6u+ a.36 ® (2)
2 S/ 2 S/ 2
Sol : ¬ ¾¾
V0
®¬¾ ¾®
V1 V2
(1)´ 3 - (2)gives
72 = –12a t t
a = –6 cm/s2 2 2
1 Average velocity for the second half distance =
100 = 2u - ´ 6´ 4
2
t t
2u = 112 u = 56cm/s v1 + v 2
V = u+ at = 56 –6 × 8 2 2 v1 + v2
t t =
\ V = 8 cm / s
+ 2
2 2

Average velocity for the first half distance = = v0 Problem : 4.18

( \ it is constant) A driver can stop his car from the red signal at a
Average velocity for total path distance of 20m when he is driving at 36 kmph and
41.25m when he is driving at 54kmph. Find his
(v1 + t 2 )
2v0 2v0 (v1 + v) reaction time.
2 = 2
=
v0 +
v1 + v2 v1 + v 2 + 2v0 s = ut + u
2a
2
é 100 ù
Problem : 4.16 ê20 = 10t + ú´ 2.25
êë 2a úû
A particle traversed along a straight line for first 225
half time with velocity V0 . For the remaining part, 41.25 = 15t +
2a
half of the distance is traversed with velocity V1 and 3.75 = 7.5t
other half distance with veloctity V2. Find the mean
velocity of the pariticle for the total journey. t = 0.5s
t t Problem : 4.19
2 2 A car starts from rest and moves with uniform
Sol. v0 x x acceleration 'a'. At the same instant from the same
2 2 point a bike crosses with a uniform velocity 'u'.When
For the first half time average velocity = V0 and where will they meet ? what is the velocity of car
2v1 v 2 with respect to the bike at the time of meeting?
For the second half time average velocity = v + v 1 2
1 2 Sol. Sr = ur t + ar t
2
2v1v 2 1 2
v0 +
Average velocity for total journey = v1 + v 2 0= ut – at
2
2 2u
t=
v 0 (v1 + v 2 )+ 2v1 v 2 a
average velocity = 2u 2u 2 2u 2
Sbike = u.t = u. = Sbike =
2 (v1 + v 2 ) a a a
Vcar =at = 2u
Problem : 4.17 Vcar w.r. t. bike at the time of meeting = 2u –u
A car is moving with a velocity of 20 m/s. The driver Vcb = u
sees a stationary truck ahead at a distance of 100 m.
Problem : 4.20
After some reaction time D t the brakes are applied
producing a retardation of 4 m/s2. What is the maxi- A bus starts moving with acceleration 2m/s2. A boy
mum reaction time to avoid collision ? 96m behind the bus simultaneously starts running
Sol. The car before coming to rest with a constant velocity of 20m/s. After what time he
will be able catch the bus?
v 2 = u 2 + 2as covers distance s u= 0
u = 20m / s a = 2m / s2
2
\ 0 = 20 - 2´ 4s
Sol.
Boy
400 Bus
\ s= = 50 m
8 ¬¾
96m
¾® ¬ ¾
S
¾
® bus
The car covers 50 m 1
S r = ur t + ar t 2
To avoid the clash, the remaining distance 100 - 50 2
= 50 m must be covered by the car with uniform 1
96 = 20t - 2t 2
velocity 20 m/s during the reaction time D t 2
2
50 50 t - 20t + 96 = 0
\ = 20 \ Dt = = 2.5 s
Dt 20 on solving t = 8S and t = 12S
\ The maximum reaction time D t = 2.5 s He can catch the bus at two instants 8s and 12s. After
12 seconds the bus will always be ahead of the boy

Problem : 4.21 BC 5 2
t=  =
vBA 20 2 (BC = AC = 5 2 )
Two bodies start moving in the same straight line at
the same instant of time from the same origin. The 1
hr = 15 minutes
first body moves with a constant velocity of 40 m/s , 4
and the second starts from rest with a constant Problem : 4.23
acceleration of 4 m/s2 . Find the time that elapses
before the second. catches the first body. Find also Two trains one travelling at 54 kmph and the other
the greatest distance between then prior to it and the at 72 kmph are headed towards one another along a
time at which this occurs. 1
straight track. When they are km apart, both
2
Sol. When the second body catches the first , the distance drivers simultaneously see the other train and apply
travelled by each is the same. their brakes. If each train is decelerated at the rate of
1 ms-2, will there be collision ?
1
\ 40t = (4)t 2 or t = 20 S Sol. Distance travelled by the first train before coming
2
to rest
Now , the distance s between the two bodies at any
2
1 2 u2 æ
çç72 ´ 5 ÷
ö 400
time t is s = ut - at s1 = = ÷ / 2´ 1 = = 200 m
2 2a çè 18 ø÷ 2
ds 225
For s to be maximum, = 0 or u - at = 0 = = 122.5 m
dt 2
u 40 Distance travelled by the second train before
or t = = = 10s coming to rest
a u
2
Maximum Distance æ 5 ö÷ 400
s2 = çç72´ ÷ / 2´ 1 = = 200 m
1 2
çè 18 ø÷ 2
= 40´ 10 - ´ 4´ (10) = 400 - 200 = 200 m
2 Total distance travelled by the two trains before
Problem : 4.22 coming to rest = s1 + s2 =122.5 + 200 = 322.5 m
Because the initial distance of separation is 500 m
Two ships A and B are 10 km a part on a line running
which is greater than 322.5 m, there will be no
south to north . Ship A farthar north is streaming
west at 20 km/hr and ship B is streaming north at 20 collision between the trains.
km/hr . What is their distance of closest approach Problem : 4.24
and how long do they take to reach it ? In a car race, car A takes time t less than car B and
Sol. Ships A and B are moving with same speed 20 km/hr passes the finishing point with a velocity v more than
in the directions shown in figure . It is a two the velocity with which car B passes the point.
dimensinal , two body problem with zero accelera- Assuming that the cars start from rest and travel with
tion N
v
 
vA
E constant accelerations a1 and a2, show that  a1a 2.
vBA
Let us find 
vB t
   B Sol. Let s be the distance covered by each car. Let the times
vBA = vB - vA AB=10km
taken by the two cars to complete the journey be t1
 and t2, and their velocities at the finishing point be v1
here , vBA = (20)2 + (20)2 = 20 2km / hr
and v2 respectively.
 According to the given problem,
i.e vBA is 20 2 km/hr at an angle of 450 from east
v1  v2  v and t 2  t 1  t
towards north . Thus , the given problem can be sim-

plified as A is rest and B is moving with vBA in the v v1  v2 2a1s  2a 2 s
Now,  =
direction shown in t t 2  t1 2s 2s

Therefore , the mininum distance between the two is a2 a1
a1  a 2
S min = AC = AB sin 450 = 5 2km A
= 1 1
C
 
vBA
a2 a1
B
and the desired time is V
\ = a1a 2
t

*Problem : 4.25 (b) Displacement = area under the v-t graph

A particle moving along a straight line with initial = area of  OAB
velocity u and acceleration a continues its motion for
1 1
n seconds. What is the distance covered by it in the last   base  height   t vmax
2 2
nth second ?
1 2 1   t 
Hint. S = ut + at = t 
2 2    
1 2
Displacement in n seconds = un + an 1   t 2 
2
= 2    
Displacement in (n –1) seconds  
1
= u ( n – 1) + a(n–1)2 Problem : 4.27
2
Displacement in n th second = Displacement in n Figure shows the motion of a particle along a straight
seconds – displacement in (n – 1) seconds. line. Find the average velocity of the particle during
the intervals
æ 1ö
\ S n = u + a ççn - ÷ ÷. (a) A to E; (b) B to E; (c) C to E;
çè 2 ø÷
(d) D to E; (e) C to D.
PROBLEMS BASED ON GRAPHS
Problem : 4.26 Sol. x cm
A bus accelerates from rest at a constant rate  for C
D
12
some time, after which it decelerates at a constant E
rate  to come to rest. If the total time elapsed is t 8
seconds then, evaluate. B
4
(a) the maximum velocity achieved and
A
(b) the total distance travelled graphically. O 3 5 8 10
t (seconds)
Sol. (a) Let t1 be the time of acceleration and t2 that of
deceleration of the bus.
(a) As the particle moves from A to E, A is the initial
The total time is t = t1 + t2. point and E is the final point.
Let vmax be the maximum velocity. The slope of the line drawn from A to E
As the acceleration and deceleration are constants the
velocity time graph is a straight line as shown in the i.e., x gives the average velocity during that interval
figure.with +ve slope for acceleration and -ve slope t
for deceleration. of time.
From the graph, The displacement x is
the slope of the line OA gives the acceleration  . xE – xA = 10 cm – 0 cm = +10 cm
vmax vmax
 = slope of the line OA = t  t1   The time interval t EA = tE - tA = 10s.
1
the slope of AB gives the deceleration   During this interval average velocity
vmax v x 10cm
  = slope of AB =  t2  max v   1cms 1
t2  t 10 s
vmax vmax (b) During the interval B to E, the displacement
t = t1 + t2 = 
  v x  x E  x B  10cm  4cm  6cm and
    vmax A
t  t E  t B  10s  3s  7s.
t  vmax  
   D x 6cm
 Average velocity v= =
æ a b ö÷ B
Dt 7s
t
\ vmax = çç ÷t O t1 t2
çèa + b ø÷ = +0.857 cms–1 = 0.86cms–1

(c) During the interval C to E,the displacement The equation of motion for this body which gives
x  x E  xC  10cm  12cm  2cm and 1 2
variation of displacement with time is S = 7t - 0.64t
2
D t = tE - tC = 10s - 5s = 5s =7t - 0.32t2.
D x - 2cm
\ v= = = - 0.4cms- 1 *Problem : 4.29
Dt 5s
(d) During the interval D to E, the displacement The graphs in (i) and (ii) show the S – t graph and
V – t graph of a body.
x  x E  x D  10cm  12cm  2cm
Are the motions shown in the graphs represented by
and the time interval OAB the same ? explain
t  t E  t D  10 s  8s  2 s
x cm A
x  2cm
v    1cms 1 10
t 2s
(e) During the interval C to D,the displacement
x  x D  xC  12cm  12cm  0
5
and the time interval t  t D  t C  8s  5 s  3s (i)
x 0m
1 B
 The average velocity v  t  3s  0 ms O 5 10
f, seconds
(The particle has reached the same position during
these 3s. The average velocity is zero because the
displacement is zero). (ii) v cm/s
5
*Problem : 4.28 A
Velocity–time graph for the motion of a certain body
is shown in Fig. Explain the nature of this motion.
Find the initial velocity and acceleration and write O
the equation for the variation of displacement with B 5 10 t, seconds
i
time. What happens to the moving body at point B ?
Sol. The motion shown by the two graphs are not same.
How does the body move after this moment ?
Sol. i) In the given s – t graph OA, is a uniform retardation
V m/s motion.
Here,
A displacement = (average velocity) x (time )

5 æu + 0 ö÷
\ 10 = çççè 2 ø÷
÷× 4
B ts
0 15 \ u = 5ms–1
5 10
C using v2 – u2 = 2as
The velocity – time graph is a straight line with–ve
0 – 52 = 2a(10)
slope. The motion is uniformly retarding upto point B
a = – 1.25 ms– 2
and there after uniformly accelerated upto C.
At point B the body stops and then its direction of ii) In the given V – t graph, OA is a uniform retardation
velocity reversed. of motion
OA - 4 ms- 1
The initial velocity at point A is v0 = 7 ms-1. a = slope of the line = = = - 1ms- 2
OB 4s
v f - vo 0 - 7ms - 1 - 7 - 2 Thus the two graphs even though represent uniform
\ a=
Dt
=
11s
=
11
ms = 0.64ms2
retardation motions, the magnitudes are not equal.

*Problem : 4.30 Vav 

S  2S  5S
2S 2S 10S
The acceleration - displacement graph of a particle  
Vmax Vmax Vmax
moving in a straight line is given as in the fig. The
initial velocity of the particle is zero. Find the velocity Vav 8S 4
  (OR)
of the particle when displacement of the particle is Vmax 14S 7
s = 12m. Vav Total displacement
= ¾¾ ¾¾ ¾¾¾ ¾ ¾ ¾¾ ö®
a(ms–2) Vmax
æ
çtotal displacement ö÷
2ççduring acceleration÷
æDisplacement
÷+ ççDuring uniform÷
÷ çç
÷
÷
÷
ç ÷ ÷
èçandretardaton ø÷ èçvelocity ø÷
Vav 8S 8 4
4 \   
Vmax 2  S  5S  2S 14 7
2
Problem : 4.32
2 8 10 12 S(m) Figure given here shows the displacement time graph
for a particle. Is it practically possible ? Explain.
Sol. From the equation v2 - u2 = 2as
v2  u 2
as = = area under a-s graph
Displacement
2
time
initial velocity u = 0;
v2
 as  = area under a-s graph.
2
Sol. From the graph, it is evident that, at any instant of
V  2  area under a  s graph  time the particle possesses two displacements, which
is impossible.
1 1 1 
   2  2   6  2   2  4  2   2  4  2  2  24  4 3ms 1
2 2 2  Problem : 4.33
Figure given here shows the variation of velocity of a
*Problem : 4.31
particle with time.
A body starts from rest and travels a distance S with
uniform acceleration, then moves uniformly a
distance 2S and finally comes to rest after moving
further 5S under uniform retardation. Find the ratio v ms-1
of average velocity to maximum velocity :
8
Sol. Graphically : Area of (V–t) curve represent 4
displacement 2
0 t
2 4 7
Find the following :
Vmax Vmax
i) Displacement during the time intervals.
V a) 0 to 2 sec. b) 2 to 4 sec.and c) 4 to 7 sec
t ii) Accelerations at
t1 t2 t3 a) t = 1 sec, b) t = 3 sec. andc) t = 6 sec.
1 2S iii) Average acceleration
S= V t or t1 = V
2 max 1 max a) between t = 0 to t = 4 sec.
2S b) between t = 0 to t = 7 sec.
2S = Vmax t2 or t2 = V
max iv) Average velocity during the motion.
1 10S Hint.
5S = Vmaxt3 or t3 = V
2 max i)displacement = Area enclosed between v – t graph
Total displacement and time axis.
Vav 
Total time ii) Acceleration = slope of v – t curve

Total change in velocity ds

iii) Average acceleration = Sol : v= = constant in the part bc \ a = 0
Total time dt
S
Total displacement
iv) Average velocity = c
Total time
b d
Ans. (i) a) 8m b) 16 m c) 12m
l
–2 –2
ii) a) 4ma b)0 c ) – 2/3ms 0 t
iii) a) 2m/s2 b) 0
Þ F = 0 in the part bc
36 - 1
iv) ms
7 Note 18 : The acceleration, time graph is as shown
Problem : 4.34 a
The velocity time graph of a moving object is given

in the figure. Find the maximum acceleration of the
body and distance travelled by the body in the interval
of time during which this acceleration exists.
t
t1 t2
Velocity (in m/s)
80 The corresponding v–t graph will be

60
40 v
20
0 10 20 30 40 50 60 70 80
Time (in)
Sol. Acceleration is maximum when slope is
80 - 20 t
= 6m / s2 t1 t2
maximum a max =
40 - 30
1 2 2 Problem : 4.37
S = 20 m/s × 10s + x 6m/s × 100 s = 500 m
2 Three particles start from the origin at the same time,
Problem : 4.35 one with a velocity v1 along x - axis , the second
The velocity–time graph of a body moving in a along the y - axis with a velocity v2 and the third
along x = y line. The velocity of the third so that the
straight line is shown in Fig. Find the displacement
three may always lie on the same line is
and distance travelled by the body in 10 sec.
Sol. Let time interval be chosen as 1 second.
u(m/s)
A PA OA vx
= =
20 D PB 0 B v y
10 So , P ( x ,y) divides AB in the ratio vx : vy
0 t(s)
6 8 10 vx ´ 0 + v y ´ vx vx v y
2 4
10
x= =
vx + v y ux + vy
20 C
(0,vy) B
Hint. The area enclosed by velocity-time graph with
time axis measures the displacement travelled in the p
(x,y)
given time. vy
-2
A
Ans. S = 50m
0 vx
Problem : 4.36
vxvy + vy ´ 0 vxv y
S –t group of a particle moving an a straight line is y= =
vx + v y vx + v y
as shown. On which part the force acting is zero.

v= x2 + y 2 v
O - V = b t3; t3 =
b
vx v y
= 2 O 2 - V 2 = - 2b S3 v
vx + vy
Now, repalce vx by v1 and vy by v2 v2 1 2
Problem : 4.38 S3 = = b t3
2b 2
The displacement - time graphs of two particles P t1 t 2 t 3 t
and Q are as shown in the figure. The ratio of their S S - (S1 + S3 )
t2 = 2 =
velocities Vp and VQ will be V V
v2 æ ö
çç 1 + 1 ÷
S- ÷
2 èa b ø÷
ç
=
V
60 0
t = t1 + t 2 + t 3
30 0
v s v v v
= + - - +
a v 2a 2b b
Sol. The velocity of a particle is equal to the slope of time
- dispalcement straight line. s væ a + b ö÷
t= + çç ÷
1 v 2 çè a b ø÷
VP tan 300 3 = 1: 3
= = s 2 = s - (s1 + s 2 )
VQ tan 600 3
æv 2 v 2 ö÷
= s - çç + ÷
Problem : 4.39 çè 2a 2a ø÷÷
The a - t graph is shown in the figure. The maximum
velocity attained by the body will be
v2
= s-
a
10
= s - (vt - s)
a(ms -2)
s 2 = 2s - vt
0
t(s) 11 [S2= distance travelled with constant velocity]
Sol : Maximum velocity = at PROBLEMS BASED ON CALCULUS
=Area between v-t graph and t-axis
Problem : 4.41
1
  11 10  55 m A point moves the plane x - y according to the law
2
Problem : 4.40 x = k sin  t and y = k(1–cos t ) where k and  are
positive constants. Find the distance s traversed by the
A car travels starting form rest with constant particle during time t.
acceleration a and reaches a maximum velocity V..
dx
It travels with maximum velocity for some time and Sol.  v x  k  cos t
dt
retards uniformly at the rate of b and comes to rest.
dy
If s is the total distance and t is the total time of and  v y  k  sin t
dt
journey then t =
t1 t3 t2
Now, speed v = v 2
x 
 v 2y  k  cons tant
Sol : VS aS
V Sb
 s = vt = kt .
1 2
3
v Problem : 4.42
V = 0 + a t,; t1 =
a The coordinates of a body moving in a plane at any
2 2
1 v instant of time t are x = a t 2 and y = b t .
S1 = a t12 =
2 2a The velocity of the body is

dx Problem : 4.45
Sol. x = a t 2 Þ vx = = 2a t
dt Acceleration of a particle is varying according to
dy the law a = – ky. Find the velocity as a function of y,
y = b t 2 . Þ vy = = 2b t
and intial velocity V0
dt
dv dv dy
\ velocity v = vx2 + vy2 Þ
2 2
(2a t ) + (2b t ) Sol. a = =
dt dy dt
V y
= 2t a 2 + b 2
ò v dv = ò - kydy Þ V02 - V 2 = ky
Problem : 4.43 V0 0
The motion of a particle along a straight line is de- V= V02 - ky 2

scribed by the function s = 6 + 4t 2 - t 4 in SI units.
Find the velocity, acceleration, at t=2s, and the aver- Problem : 4.46
age velocity during 3rd second. The velocity of a particle moving in the positive
Sol. 2
s = 6 + 4t - t 4 direction of the X-axis varies as V = K S where K is a
positive constant. Draw V – t graph.
ds
Velocity = = 8t - 4t 3 when t = 2 Sol. V=K s
dt
S t
Velocity = 8 × 2 -4 × 23 dS dS
= K S \   K dt
dt 0 S 0
Velocity = 16 m/s
d 2s \ 2 S = Kt and S = 1 K2 t2
Acceleration a = = 8 - 12t 2 when t=2 4
dt 2
dS 1 2 1
acc = 8-12 × 22 = -40 Þ V= = K 2t = K 2t
dt 4 2
acc = -40 m/s2 \ Vµt
displacement in 2 seconds
\ The V – t graph is a striaght line passing through
s1 = 6 + 4. 22 -24 = 6 m the origin
displacement in 3 seconds
Problem : 4.47
s2 = 6+ 4. 32 - 34 = -39 m
In the arrangement shown in figure the ends of an
displacement during 3rd second
inextinsible string move downwards with uniform
= s2 - s1 = -39 - 6 = -45 m speed u. Pulleys A and B are fixed. Find the speed
\ Average velocity during 3rd second with which the mass M moves upwards.
± 45
= = - 45 m / s A
1 B
-ve sign indicates that the body is moving in opposite  
direction to the initial direction of motion.
O
Problem : 4.44
o M o
A particle moves according to the equation
Sol. Suppose the distance of point O from the ceiling is y and
t = x + 3 , where will be the particle come to the
the distance of point O from each pulley is x and the
rest for the first time
distance between the two pulleys is  .
Sol. x = (t – 3) 2 l/2
2
x= t2 – 6t + 9 l y
x2  y 2   x
dx 4
v= = 2t - 6
dt Differentiating the above eqation w.r.t to “ t ”
0 = 2t – 6
ædx ö÷ ædy ö÷ 1 æd  ö÷
2 x çç ÷ = 2 y çç ÷ + 2 ç ÷
Þ t = 3s çè dt ø÷ èç dt ø÷ 4 èçç dt ø÷

dx dy d
But, u , v= and Problem : 4.50
dt dt dt
x A point moves the plane x – y according to the law
\ - xu = yv and v = – u
y = u sec  x = k sin  t and y = k (1–cos  t ) where k and  are
 Velocity of mass = v = u secθ (upwards) positive constants. Find the distance s traversed by the
Problem : 4.48 particle during time t.
Figure shows a rod of length  resting on a wall and the dx
Sol.  v x  k  cos t
floor. Its lower end A is pulled towards left with a constant dt
velocity v. Find the velocity of the other end B downward
dy
when the rod makes an angle  with the horizontal. and  v y  k  sin t
B
dt
 Now, speed v = v 2
x 
 v 2y  k  constant
 s = vt = kt .
V A 
Problem : 4.51
Sol. In such type of problems, when velocity of one part of t = ax2 + bx find acceleration ?
a body is given and that of other is required, we first (a, b are constants)
find the relation between the two displacements, then Sol. t = ax2 + bx
differentiate them with respect to time. Here if the
dt
distance from the corner to the point A is x and up to B = 2ax + b
is y. Then dx
dx dy dx 1
v= & vB=– v= =
dt dt dt 2a + b
(–sign denotes that yis decreasing) dv - 2a dx
=
Further, x2 + y2 =  2 dt (2ax + b)2 dt
Differentiating with respect to time t - 2a
dx dy
= = - 2av3
2x  2y 0 xv = yvB (2ax + b)3
dt dt
x Problem : 4.52
vB = (v) =v cot 
y
A point moves rectilinearly with deceleration whose
Problem : 4.49
modulus depends on the velocity v of the particle as
Acceleration of a particle at any time t is a = k v, where k is a positive constant. At the initial
 moment the velocity of the point is equal to V0. What
 
a  2tiˆ  3t 2 ˆj m / s 2 . If initially particle is at rest,
distance will it take to cover that distance?
find the velocity of the particle at time t = 2s.
Sol. Here acceleration is a function of time, i.e., Sol. Let t0 be the time in which it comes to a stop. Given
acceleration is not constant. So, we cannot apply dv
   that - = k V
v  u  at . dt
We will have to go for integration for finding velocity t0 0
dv
at any time t. ò kdt = ò-
 v
dv   v0 v0
a= Thus dv  a dt

dt  kt0 = 2 v0
v 2
   2
or  dv   adt or  
v   2tiˆ  3t2ˆj dt
\ t0 =
2
v0
0 0 0
2 k
2 3ˆ
0

=  t î  t j  = 4iˆ  8jˆ m/ s  Now to find the distance covered before stopping,
Therefore, velocity of particle at time t = 2s is dv dv ds dv
 4iˆ  8ˆj m/ s =
dt ds dt
= v
ds
dv ii) V = S + 5
But, = - k V;
dt ds
\   S  5
dv dt
\ v =- k V
ds S
dS
t
\ ò = ò dt
\ vdv = - kds S+ 5
0 0
s  S 5
0 s
2 3
2
\ loge  S  50  t , loge  =t
\ ò vdv = - ò k ds Þ s = 3k
V
0  5 
V0 V0
Problem : 4.55
Problem : 4.53 The x and y co-ordinates of a particle are
A particle moves according to the equation x = A sin (wt ) and y = A sin (wt + p / 2). Find the
dv motion of the particle
= a - b v where a and b are costants. Find
dt Sol. Given x = A sin wt
velocity as a function of time. Assume body starts
y = A sin (wt + p / 2)= A cos wt
from rest.
squaring and adding 1) and 2) we get
dv Y
Sol. = a - bv
dt
v t t  0
dv
ò a - bv = ò dt x2+y2=A2 X
0 0
 t  3 / 2 t   / 2
V
éln (a - b v )ù t  
ë û0 = t
-b
ln (a - b v )- ln a = - b t i.e . path of the particle is a circle with centre at
a - bv origin and radius A
ln = - bt At time wt = 0 x = 0 and y=A
a
and at wt = p / 2 x = A and y=0
a - bv
= e- b t at wt = p x = 0 y = –A and so on.
a
b  The motion is circular clockwise
1- v = e- b t
a Problem : 4.56
a
v = (1- e- b t ) An object is projected in X – Y plane in which velocity
b 
changes according to relation V  aiˆ  bxjˆ . Equation
of path of particle is:
Problem : 4.54
a) Hyperbolic b) Circular
The acceleration a of a particle depends on c) Elliptical d) Parabolic
displacement S as a = S + 5. It is given that initially
dx
S = 0 and V = 5 m/s. Sol. a
dt
Find relation between i) V and S ii) S and t
x = at
dV dV dS Vy = bx = b at
Sol. i) : a = = . = S+ 5
dt dS dt
dy
dV  b at
\ V = S+ 5 dt
ds
V S  dy   bat dt
\  V dV    S  5 dS bat2
5 0
y=
2
éV éS 22 ùV
ùS ba x 2 bx 2
Þ êê úú = êê + 5S úú y= 
ëê 2 úû5 êë 2 ûú0 2 a2 2a
y  x 2 i.e., parabolic.
 V = (S+5)

Problem : 4.57 4.25 EQUATIONS OF MOTION FOR FREELY

The displacement (x) of a particle varies with time FALLING BODY
asx = ae–  t + be  t where a, b, ,  are positive constant. Motion of all the bodies falling towards the Earth
Find how does the velocity of particle change with time, when air resistance is ignored is known as free fall.
Sol : x = ae–  t +be  t For a freely falling body, u = 0, a = +g
dx  (i) v  u  at  v = gt
V  aet  bet
dt
1 1
V =  bet  aet (ii) S  ut  at 2  S  gt 2
2 2
\ V increases as t increases.
(iii) v 2  u 2  2 aS  v 2  2gS
4.24 ACCELERATION DUE TO GRAVITY (G)
 1
The uniform accleration of a freely falling (iv) Sn  u  a  n   
 2 
body towards the centre of earth due to earth’s
gravitational force is called acceleration due to  1
Sn  g  n  
gravity.  2
i) It is denoted by ‘g’ 4.26 EQUATIONS OF MOTION FOR
ii) Its value is constant for all bodies at a given VERTICALLY PROJECTED BODY
place. It is independent of size, shape, material, For a body projected vertically upwards,
constitution(hollow or solid), nature of the body. If a = – g (since velocity, acceleration vectors are
air resistance is ignored, all the bodies as light as a opposite)
feather to a heavy met al sphere, dropped  (i) v  u  at  v  u  gt
simultaneously from the same height hit the floor at 1 2 1 2
the same time because all the bodies have same (ii) S  ut  at  S  ut  gt
acceleration due to gravity. 2 2
2 2 2 2
(iii) v - u = 2aS  v - u = 2  - g  S
iii) Its value changes from place to place on the
surface of the earth.  1  1
(iv) Sn  u  a  n    Sn  u  g  n  
iv) It has maximum (greatest) value at the poles of  2   2 
the earth. The value is nearly 9.83 m/s2.
4.27 MOTION PARAMETERS OF A BODY
It has minimum (least ) vlaue at the equator of
PROJECTED VERTICALLY UPWARDS
the earth. The value is nearly 9.78 m/s2.
i) Maximum height (Hmax) :
v) The average value of g on earth’s surface is 9.8 m/s2.
“For a body projected vertically upwards, the
g earth
vi)
2
On the surface of moon, g  1.67m / s  maximum vertical displacement from ground about
6 which its velocity is zero is called its maximum
On the surface of sun, g = 274 m/s2 height.”
vii) The acceleration due to gravity of a body is Expression :
always directed downwards towards the centre of Let a body be projected vertically upwards with
the earth, whether a body is projected upwards or
initial velocity u.
downwards.
We know that, v2 – u2 = 2as
viii) When a body is falling towards the earth, its
velocity increases, g is positve. here a = – g, s = Hmax, v=0
2
ix) When a body is projected upwards, its velocity  0 – u = 2 (– g) Hmax
decreases, g is negative.  – u2 = – 2gHmax
x) The acceleration due to gravity at the centre of u2
 H max 
earth is zero. 2g

(ii) Time of ascent (ta) : iv) Time of flight ( tf ) :

“For a body projected upwards the time to reach “For a body projected vertically upwards the
the maximum height is called time of ascent” sum of time of ascent and time of descent is called
Expression : time of flight (tf)” It is the total time for which the
Let a body be projected vertically upwards with body remains in air
initial velocity u. Time of flight = Time of ascent + Time of descent
We know that, v = u + at t f = ta + td
Here a = – g, t = ta, v = 0 u u 2u
 tf   
 0  u  gta  u  gta g g g
2u
u  tf 
 ta  g
g
v) Velocity of the body on reaching the point
(iii) Time of descent (td) :
of projection (Vstriking)
“For a body projected upwards the time to travel
Let a body be projected vertically upwards with
from maximum height to the point of projection on initial velocity u.
ground is called time of descent”
The body reaches the point of projection once
Expression : Let a body be projected verti- again after the time of flight (tf)
cally upwards with velocity u. We know that, V = u + at
Step 1) : For upward motion Here,
initial 2u
velocity  0 a = – g, t = tf = g , V = Vstriking
0  u 2  2   g  H max
u2
S  H max  2u 
 H max   V striking = u - g    u
2g  g 
u
 V striking =  u
Step 2) : For downward motion
The body reaches the point of projection with
1 2 the same speed of projection but in opposite direction.
S = (initial velocity) t + gt
2
Here, Note 4.19 : For a body projected vertically upwards s-t , v-
t , a-t graphs are as follows :
initial velocity = 0, a   g , t  td , S  H max v a
s
t
1 u2
 H max  0  gtd2
2 2g t
2 -g
u 1
  gtd2 u 2u
t
2g 2 g g
u
u2  td  Note 4.20 : For a freely falling body s-t , v-t , a-t g
 t d2  g
g2 raphs are as follows :
Note 4.19 : a
For a body projected vertically upwards, s v

u tanq= g
Time of ascent (ta) = Time of descent (td)= q
g t t t

Note 4.21: In the case of a vertically projected body, Note 4.29:

velocity is maximum at projection point and mini- Distance covered by a body projected vertically
mum (zero) at highest point up in the last one second of its upward journey =
Distance covered by it in the 1st second of its downward
Note 4.22 : Velocity goes on decreasing . g
journey =
Note 4.23: Velocity at any point during the upward 2
journey = velocity at the same point during the Sol. For a body falling downwards , we know that
downward journey (numerically) 1
s = gt 2
Thus projection velocity = Striking velocity (nu- 2
g
merically) . Substituting t=1 gives us s = (clearly , the
st
2
Note 4.24: Change in velocity in the entire journey distance travelled in1 second in the downward
journey is same as that of last second of upward
= 2u {( D v = v f - vi = u - (- u )= 2u }
journey )
Note 4.25: Similarly Change in momentum. during
Note 4.30:
the entire journey = 2mu
Time taken by vertically projected up body to
Note : 4.26 3 t
u reach th of maximum height = a
Velocity at half maximum height = 4 2
2 Sol. we know that , for a body projected vertically
Sol. using the equation v2-u2=2as, we have a=-g , u 3
up v=u-gt Þ = u - gt (since at th of maximum
1 1 u2 2 u 4
s= H= , we get height , velocity = )
2 2 2g 2
u2 - u2 u2 u
u u 1
v 2 - u 2 = 2 (- g )´ Þ v2 - u 2 = Þ v2 = Þ v= Þ gt = Þ t = Þ t = ´ ta
4g 2 2 2 2 2g 2
Note 4.27: Note 4.31:
3 u
Velocity at th of maximum height = A body projected vertically up takes a time
4 2
æ 1 ö÷
Sol. using the equation v2-u2=2as, we have a=-g , t = ta çç1- ÷ to reach half of maximum height .
3 3 u2
èç 2 ø÷
s= H= , we get Solution : we know that , for a body projected
4 4 2g u
2 2
v - u = 2 (- g )´
3u 2
Þ v2 - u 2 =
- 3u 2
Þ v2 =
u2
Þ v=
u vertically up v=u-gt . Þ = u - gt (since at half of
8g 4 4 2 2 u
Note 4.28 : Distance covered by a body projected maximum height , velocity = )
2
vertically up in the 1st second of its upward journey u uæ 1 ö÷ æ 1 ö÷
u- = gt Þ t = çç1- ÷Þ t = ta çç1- ÷
= Distance covered by it in the last second of its ÷
g 2 g èç 2ø èç 2 ø÷
entire journey = u - Note : 4.32
2
Sol. we know that for a body projected vertically When air resistance is taken into account
æ 1ö
up, sn = u - g çççèn - ø÷
÷. i) Time of ascent is less than that in vacuum
2÷
g ii) Time of ascent is less than time of descent
Substituting n=1 gives s1 = u - (clearly , the
2 iii) The speed of the body when it reaches the
distance travelled in last second is same as that of 1st
point of projection is less than the speed of
second ) projection
O h
u
a) height at which they meet is t =
u
R Sol. Let the two meet after a time 't' seconds then
the in this the distance covered by both must
mg be equal to height of true
mg R
v
i.e , S1+S2=h
u
F = mg + R F = mg – R 1 2 1
Þ gt + ut gt 2 = h Þ ut = h
R 2 2
a = g+ a1 = g - R m h
m Þ t=
0 = u – ata v= 0 + a1td u
b) the time after which their velocities are
u æR ö u
ta = v = g - çç ÷ equal is t =
g+ R ÷t d
çè m ø÷ 2g
m
Sol : Let the velocities be equal after a time 't'
1æ ö 2 1 æ R ö÷ 2
ççg + R ÷
h= ÷t a = çççg - ÷t d Þ v1 = v2
2 èç m ø÷ 2è m ø÷ u
Þ gt = u - gt Þ u Þ t =
2g
ta g- R c) Ratio of distances covered when the
= m= v
td g+ R u magnitudes of their velocities are equal is
m
S1 : S2 = 1: 3
v g- R g- R
\ = m ´ td m Sol : From above , velocities are equal after a time
u g+ R t g+ R u
m d m t= in this time
For dropped bodies 2g
2
i) Same resistance force R 1 æ ç u ÷ö 1 u2
S1 = g ç ÷ = ´ g ´
2 çè 2 g ÷
÷
ø 2 4g 2
Þ a= g- R/m
u2
If m is more a is more Þ S1 =
8g
Þ heavier body falls first 2
1 æ u ö÷ 1 æ u ö÷
ii) If R is proportional to m then acceleration is Þ S 2 = ut - gt 2 = u çç ÷- gç ÷
same for both 2 ÷ 2 èçç 2 g ø÷
èç 2 g ø÷ ÷
Þ both the balls fall simulateneously in the u 2 u 2 3u 2

same time = - = Þ S1 : S 2 = 1 : 3
2 g 8 g 8g
iii) If m is same R is less for smaller body; a =
g–R/m is more for smaller body Application : 4.12
Þ smaller body falls first Three bodies are projected from towers of
same height as shown. 1st one is projected
Application : 4.11 1 vertically up with a velocity 'u'. The second one
u=0
Body 1 is released from the top is thrown down vertically with the same velocity
of a tower . At the same instant, h and the third one is dropped as a freely falling
body 2 is projected vertically up 4
body. If t1, t2, t3 are the times taken by them to
as shown then 2 reach ground, then,

1 i,e . at P, g = 0 Þ v = cons tan t

a) velocity of projection is u = g (t1 - t2 )
2 Step a] Distance converd in t second is
1 2
u=u u=0 gt = x
2
h h h remaining distance PQ is to be convered by
the body with constant velocity for which have
1 2 3 SPQ = VP x t'
Sol. Clearly the extra time taken by the 2nd body Step b] Velocity at P is VP = gt
is equal to the time of flight of 1st body Step c] SPQ = VP x t' (t'=time taken to cover the
2u 1 distance PQ)
i.e, t1 - t2 = Þ u = g (t1 - t2 )
g 2 SPQ H- x
\ t' = =
1 VP gt
b) height of the tower is h = gt1t2
2 1 2
H- gt
Sol. We know that , for a vertically projected up 2 H t
\ t= = -
body gt gt 2
1 iv) total time of fall is
s = ut - gt 2
2 ïì H t ïü
1
Þ h = ut - gt 2 [similar to ax2+bx+c=0] T = t + t ' = ïí t + - ïý
2 ïîï gt 2 ïþï
The product of these far roots of their Application : 4.14
chapton gives
A particle is projected vertically upwards
1 and it reaches the maximum height H in time T
h= gt 1t 2 seconds. The height of the particle at any time t
2
will be.
c) The time taken free fall in the 3rd case
é 1 2ù
2h êH - g (t - T ) ú
is given by t = t1t2 t = ëê 2 ûú
g
Sol. From v=u+gt
in above , the form that
o = u-gt
1 2u u = gT ..............i
h= gt1t2 Þ t1t2 =
2 g u2 1 2
H= = gT .......ii
2h 2g 2
t free = = t1t2
g Let h be the distance travelled in time t.
Application : 4.13 1 2
Then h = ut - gt
A body falls freely from a height 'H'. After 2
1 2
t seconds of fall, gravity casses to act. Find the h = gTt - gt ........iii
2
time of flight O U = 0
Subtracting (ii) from (iii)
Sol. Let the total time taken be T 1 1
h - H = gTt - gt 2 - gT 2
x 2 2
Let gravity cease at P g g 2
H P - (- 2Tt + t 2 + T 2 )= - (T - t )
2 2
1 2
Q \ h = H - g (T - t )
2
Application : 4.15 Application : 4.16

Roket is fired vertically up with an After falling for t1 sec, a stone hits a hori-
acceleration a. Fuel is exhausted after t sec. zontal glass plate, where it looses 50% of its ve-
Maximum height it can reach is given by locity. It then takes t2 sec to reach ground. Find
é1 æ ö÷ù the Height of the glass plate above the ground.
ê at2 çç1 + a ÷ ú
ê2
ë èç g ø÷
÷ú
û Sol. Let P be the upper edge and Q be the lower
Sol. Step I :- Distance covered by the rocket in t edge of the glass plate, as shown.
sec. (till fuel is exhausted) Step -I :- Velocity of the stone at P, Vp = gt1
1 gt
S1 = at 2 Step -II:- Velocity of the stone at Q, vQ = 1
2 2
Step II : Velocity at the end of t sec, ( it looses 50% of velocity)
v=u+at = 0+at = at ( \ u = 0 for a Step - III: from Q to R, the stone travels as a
rocket) vertically thrown down body with an initial velocity
u=0 O
Step III:- Further distance it can go up after t t1 sec

sec.
(after fuel is exhausted) is given by
P
gt
(velocity )
2
= 1 Q
stopping distance S 2 = = 2
t2 sce
2´ retardation
2 2 2
v 2 (at ) at R
= = Step -IV : Hence using the equation,
2g 2g 2g
1 2
Step -IV = Maximum height the rocket car s=ut+ at
2
reach is , H= S1 +S2 gt
Where S = QR ; u = vQ = 1 ; a = + g
1 2 1 a 2t 2 1 2 æ aö 2
H= at + = at çç1 + ÷÷ t = t2, we have,
2 2 g 2 çè g ø÷
÷
height of the glass plate above the ground =
Note : 4.33 gt 1
QR = 1 t2 + gt22 . From this equation, t2
For a Vertically thrown up body, maximum 2 2
can be found.
1 2
height H = gT where 'T' is the time of flight Application : 4.17
8
2u A frictionless wire is fixed between A and
Sol. We know that time of flight T = B inside a sphere of radius R . A small ball slips
g
along the wire.Find the time taken by the ball to
gT
Þ u= slip from A to B
2 A
A g
\ Maximum height
2 R
ægT ö÷ q O
g cos q

ç ÷ B
u 2 ççè 2 ø÷ 1 2 R O
H= = = gT 90 0
R
2g 2g 8 B
Similarly the same formula is applicable even 1 2

Sol. S = at
in the case of a projectile. 2
1 c) A body dropped from the top tower takes

i.e. AB = (g cos q)t 2 = 2R cos q
2 times t = t1t2 to reach the ground
R 1 1 2
\ t= 2 H  g  t1t2  = g t
g 2 2
4.28 VERTICAL PROJECTION OF AN \ t = t1t2
OBJECT FROM A TOWER Hint.
(EXPRESSION FOR HEIGHT OF TOWER) For the body projected vertically upwards,
Consider a tower of height H. Suppose a body 1
is projected upwards vertically with initial velocity H  gt12  ut1 ------- (1)
2
u from the top of tower. Suppose it reaches to a dis- For the body projected vertically downwards
placement x above the tower and there after reaches
1 2
the foot of the tower. Let t be the total time of travel. H  ut2  gt2 -------- (2)
2
Now, considering the total path of the body,
Solve (1), (2)
the motion parameters are as follows.
Initial velocity of the body = u Application : 4.19
A body projected vertically upwards from
Net displacement of body = S = +x–x–H= –H
ground is at the same height h from the ground
Time of travel = t at two instants of time t 1 and t 2 (both being
1 measured from the instant of projection) Now
we know that, S  u t  a t 2
2 1
Here, a) h = g  t 1t 2 
x 2
a = - g, S = - H, u = u, t = t x 1
b) Velocity of projection = u = g  t 1 + t 2 
1 u 2
 H  ut  gt 2 1 2
2 T
O
c) H max = g  t 1 + t 2 
8
1 H W H
 H  gt 2  ut E d) A body dropped from height h takes time
2 R
Note : 4.34 t1t2 to reach the ground
This is a quadratic equation in time. Problem : 4.58
Comparing with the standard quadratic A body is projected vertically up with velocity u from a
equation
tower. It reaches the ground with velocity nu. The height
ax 2 + bx + c = 0 u2 2
u  u 2 + 2gH of the tower is H =
2g
(n - 1)
we get t=
g Sol. v2–u2=2as
Application : 4.18 Here u = u, v = nu, a = – g, s = – H
A particle projected vertically up from the 2
top of a tower takes t 1 seconds to reach the (nu ) - u 2 = 2 gH u2 2
\ H= (n - 1)
ground. Another particle thrown downwards (n 2
- 1)u = 2 gH
2
2g
with the same velocity from the same point takes
Problem : 4.59
t2 seconds to reach the ground. Then
g Two bodies begin to fall freely from the same height.
a) Velocity of projection = u   t1  t2  The second body begins to fall ‘ t ’ s after the first. After
2
1 what time from the begining of first body dose the
b) Height of tower = H  g  t1t2  distance between the bodies equals to  ?
2
Sol. Let the time of fall of the 1st body be t seconds . Time of Ans. Ratio of times taken to fall equal distances is
fall of second body = t –  .
Distances of free fall of the bodies in the above time
 1 0 :  
2 1 : 
3  2 :........ :  n  n 1 
intervals respectively are 2h
gt 2 g t   
2 Hint. use t =
H1  ;H2  g
2 2
1 2 Problem : 4.62
Therefore   H 1  H 2  gt  g
If a freely falling body covers half of its total distance in
2
  the last second of its journey, Find its time of fall.
t  
g 2 Sol. Suppose t is the time of free fall.
1
Problem : 4.60 h = gt 2 ....... (1)
2
One body falls freely from a point A at a height (H + h)
h 1 2
while another body is projected upwards with an initial = g (t - 1) ............ (2)
velocity V0 from point C at the same time as the first 2 2
body begins to fall. What should be the velocity V0 of Solving 1, 2
the second body so that the bodies meet at a point B at the (
t = 2+ )
2 s
height ‘h’? What is the maximum height attained by 2nd since 2 - 2 is not acceptable.
body for the given initial velocity? What is the value of
V0 if H = h ? Problem : 4.63
A
Sol. A balloon starts from rest, moves vertically upwards
with an acceleration g/8 ms-2 . A stone falls from the
H ballon after 8s from the start. Find the time taken by the
H+h stone to reach the ground (g = 9.8ms–2)
B Sol. Step–1 : To find the distance of the stone above the
ground about which it begins to fall from the balloon.
h
1
C S = ut + at 2
2
i) Suppose the two bodies meet after t seconds. here, s = h , u = 0 , a = g/8
gt 2 1 æg ö
H .....(1) h = ç ÷8 2 = 4 g
2 2 è8 ø
gt 2 Step–2 : The velocity of the balloon at this height can be
h = Vo t - ........ (2)
2 obtained from v = u + at
Solving (1) & (2)
g
V  0   8  g
g 8
Vo   H  h 
2H This becomes the intital velocity (u|) of the stone as the
stone falls from the balloon at the height h.
u2
ii) We know that H Max  \ u| = g
2g
1 2
2 2 2 Step-3 : For the total motion of the stone h = gt - u |t
V (H + h) g (H + h) 2
H max = o = = ( Here H max > h)
2g 2g 2H 4H Here, h = 4g , u| = g, t = time of travel of stone.
1
2
iii) When H = h, we get Vo= 2gh . \ - 4 g = gt - 2 gt
Problem : 4.61 \ t 2 - 2t - 8 = 0
solving for ‘t’ we get t=4 and – 2s. Ignoring negative
For a freely falling body, Find the ratio of the times
value of time, t=4 s
taken to fall successive equal distances.


A rocket is fired upwards vertically with a net Ball A is dropped from the top of a building and at the
acceleration of 4m/s2 and initial velocity zero. After 5 same instant that a ball B is thrown vertically upward
seconds its fuel is finished and it decelerates with g. At from the ground. When the balls collide, they are moving
the highest point its velocity becomes zero. There after in opposite directions and the speed of A is twice the
it accelerates downwards with acceleration g and return speed of B. At what fraction of the height of the building
back to ground. did the collision occur ?
i) Plot velocity – time graph for complete journey Sol. Given VA = 2VB
ii) Displacement–time graph for the complete journey. Let h1 and h 2 are the distances travelled by the two balls
( Take g = 10 m/s2.)  2 gh1  2 u 2  2 gh2
Sol. Stage i : To find velocity of rocket after 5 seconds
2u 2
VA = 0 + atOA = (4)(5)= 20ms - 1  h1  4h2  ......... (1)
g
Stage ii : To find further time of ascent after 5 seconds. If they meet after t seconds, for the condition V = 2V
A B
0 = 20 – g tAB 0 + gt = 2 (u – gt)
20  gt  2u  2 gt  2u  3 gt
\ t AB = = 2seconds
10  t  2u / 3g
Here, the total vertical displacement of stage i) and stage
1 2  1 
ii) is Also h1 + h 2 = gt +  ut - gt 2  = ut
1 2  2 
= area of OAB = (7)(20) = 70m.  2u  2u 2
2 \ h1  h2  ut  u   ..............(2)
v(m/s)
s(m)  3g  3g
B 2u 2 4u 2
70 Solving (1) & (2) we get h1  ; h2 
9g 9g
A
20 50 A
h1 2u 2 / 9 g 1
B C   
O 5 7 10.7
t(s) C t(s) h2 4u 2 / 7 g 2
O 5 7 10.7
Problem : 4.67
Stage –iii : If tBC is time of descent then
An object falls from a bridge which is 45 m above the
1 2 water. It falls directly into a small row – boat moving
70 = (10) tBC  tBC = 14 = 3.7s
2 with constant velocity that was 12m from the point of
Note 4.36 : tOABC  7 + 3.7 = 10.7s impact when the object was released. What was the
speed of the boat ?
Note 4.37 : SOA = area under v – t graph s
Sol. Velocity of boat = V =
1 t
= (5)(20) = 50m Here,
2
s = 12m
Problem : 4.65 t = time of fall of object from bridge
A stone is allowed to fall from the top of a tower 300 m 2h 2  45
heigh and at the same time another stone is projected =   3s
–1 g 10
vertically up from the ground with a velocity 100 ms .
Find when and where the two stones meet ? 12
\ V= = 4ms- 1
Sol. Suppose the two stones meet at a height x from ground 3
after t seconds. Problem : 4.68
1 2 1 2 Two balls are dropped to the ground from different
x = 100t - gt .....(1) 300 - x = 0 + gt .......(2)
2 2 heights. One ball is dropped 2s after the other, but both
Solve 1, 2 strike the ground at the same time 5s after the Ist is
t = 3 sec, x = 255.9m dropped.

a) What is the difference in the heights from which they 2h

were dropped ? Hint. i) t = ii) V = 2 gh
g
b) From what height was the first ball dropped? Ans. i) 3.64 s. ii) 35.69ms–1
Sol. a) For the first ball s = h1, u = 0, t = 5 s
Problem : 4.71
1 2
\ h1  0  5  2 9.8 x5 = 122.5m A helicopter is ascending vertically with a speed of 8.0
ms–1. At a height of 120 m above the earth, a packat is
For the second ball s = h2, u = 0, t = 3s dropped from a window. How much time does it take
1 1 for the package to reach the ground ?
 h2  gt 2   9.8  9  4.9  9  44.1m 1 2
2 2 Sol. Hint : h = gt - ut
Difference in heights 2
Ans. t = 5.83s
h  h2  h1  122.5  44.1  78.4 m
Problem : 4.72
b) The first ball was dropped from a height of
If an object reaches a maximum vertical height of 23.0
h1 = 122.5m m when thrown vertically upward on earth how high
Problem : 4.69 would it travel on the moon where the acceleration due
Drops of water fall at regular intervals from the roof of to gravity is about one sixth that on the earth ? Assume
a building of height H = 16m, the first drop striking the that initial velocity is the same.
ground at the same moment as the fifth drop detaches 1
from the roof. Find the distance between the successive Hint . H µ
g
drops. 5 Ans. 138m
Sol. Step - i : Time taken by the first Problem : 4.73
4
2h An elevator ascends with an upward accleration of 0.2m/
drop to touch the ground = t 
g 3 s2. At the instant its upward speed is 3m/sec a loose bolt
2 5m high from the floor drops from the celing of the
For h = 16m, t = . elevator. Find the time untill the bolt strikes the floor
g
2 and the displacement it has fallen
Time interval between two drops is
Sol. Intial velocity of bolt relative to the floor of
æ 1 ö÷
æ1 ö÷ 1 2 tyhe elevator = 0
tint erval çç ç ÷t =
÷t =
èçç 4 ø÷ 4
èç n - 1ø÷ g acceleration of bolt relative to the
1 floor of the elevator = (9.8 + 0.2) = 10ms2
Where n = number of drops.
1
Step - ii : If t is time of the descent then 5 = ´ 10´ t 2
2
Distance between first and second drops \ t = 1 second
1 2 If s is the displacement then
= S1 - S2 = gtint é4 2 - 32 ù = 7m.
2 ê
erval ë ûú 1
s = gt 2 - ut . \ s = 1.9m
Distance between second and third drops 2
1 2 Problem : 4.74
= S 2 - S3 = gtint erval éëê32 - 22 ùûú = 5m.
2 A balloon is rising vertically upwards with uniform
Distance between third and fourth drops acceleration 15.7 m/s2 .A stone is dropped from it. After
1 2 4s another stone is dropped from it. Find the distance
= S3 - S 4 = gtint erval éëê22 - 12 ùûú= 3m
2 between the two stones 6 s. after the second stone is
Distance between fourth and fivfth drops. dropped
1 2 Sol. if ‘f’ is upward acceleration of the balloon then the
= S 4 - S5 = gtint erval éëê12 - 0ùúû = 1m.
2 acceleration of the stones relative to the balloon is (f+g).
Problem : 4.70 The initial velocity of each stone with respect to the
i) How long does it take a brick to reach the ground if balloon is zero.
dropped from a height of 65m ? ii) What will be its Let s1 and s2 be the distances of the two stones from the
velocity just before it reaches the ground ? balloon after 10s and 6 s respectively. Now

1 2 1 Probelem : 4.77
s1 =  f + g 10  = 15.7 + 9.8 100 = 25.5×50
2 2 A parachutist drops freely from an aeroplane for 10
1 2 seconds before the parachute opens out. Then he de-
s2 = (25.5)(6 ) = 25.5 ´ 18
2 scends with a net retardation of 2 m/sec2. His velocity
s1 - s2 = 25.5 ´ 32 = 816m when he reaches the ground is 8 m/sec. Find the height
at which he gets out of the aeroplane ?
Problem : 4.75
Sol : Distance he falls before the parachute opens
A body falls freely from a height of 25m (g=10m/s2) after
1
2sec gravity ceases to act Find the time taken by it to is g  100  490 m
2
reach the ground?
Then his velocity = gt=98.0 m/s = u
Sol. 1) Distance covered in 2s under gravity Velocity on reaching ground = 8 = 
1 1 retardation = 2
s1  gt 2  10  2 2  20m
2 2  2  u 2  2 as
velocity at the end of 2s 2
82   98   2   2  S
V = gt = (10)2 = 20m/s.
Now at this instant gravity ceases to act 106  90
S   2385 m
 velocity by here after becomes constant. 4
The remaining distance which is 125–20=105 m is
Total distance = 2385 + 490
covered by the body with constant velocity of 20m/s.
Time taken to cover 105 m with constant velocity is =2875 m = height of aeroplane
given by, Problem : 4.78
S 105 A stone is dropped into a well and the sound of splash is
t1 =  t1   5.25 s heard after 5.3 sec. If the water is at a depth of 122.5 m
V 20
Hence total time taken to reach the ground from the ground, the velocity of sound in air is
= 2 + 5.25 = 7.25 s Sol : If t1 is the time taken by stone to reach the ground and t2
the time taken by sound to go up, then t1 + t2 = 5.33
Problem : 4.76
1 2
A solid ball of density half that of water falls freely Since s = ut + at
2
under gravity from a height of 19.6 m and then enters 1 2
water. Upto what ? 122.5 = 0t   9.8  t1
2
depth will the ball go? Howmuch time will it take to
come again to the water surface Negiect air resistance 245 2450
 t12    25
and vescosity effects in water (g = 9.8 m/s2). 9.8 98
Sol : Velocity at the surface of water  t1  5s
v  2 gh  2  9.8 19.6  19.6 m / s  t2  0.33s

Acceleration of a body of density d in the liquid me- displacement
b Velocity of sound =
dium of density is time
 d   d  122.5
g '  g 1     g 1      g   367 m / s
 db   d / 2  0.33
2 2 Problem : 4. 79
Using v -u = 2as in the water
2 A body is thrown vertically up with a velocity of
0 - (19.6) = 2(-g)h 100 m/s and another one is thrown 4 sec after the first
h=19.6 m. one. How long after the first one is thrown will they
1 2
Using s = ut + at , in the water when the ball returns to meet?
2
the surface, s = o Sol : Let them meet after t sec.
1
0  19.6   9.8  t  a   g  1 2 1 2
2 S1  100t  gt and S2  100  t  4   g  t  4 
t = 4 s. 2 2

1 2 1 2
Problem : 4.83
100t  gt  100  t  4   g  t  4 
2 2 A ball dropped from the 9th storey of a multi - storeyed
1 2 1 building reaches the ground in 3 second. In the first
 400  g  t 2   t  4   = g.4  2t  4  second of its free fall, it passes through n storeys, where
2   2
n is equal to ( Take g = 10 m s-2)
800 1
 2t  4   20, if g  10m / s 2 Sol : 9 y = ´ 10 ´ 3´ 3 or y = 5 m
4g 2
 t  12 sec 1
Again , n´ 5 = ´ 10´ 1´ 1 = 5 or n = 1
2
Problem : 4. 80 Problem : 4.84
A lead ball is dropped into a lake from a diving board A stone is dropped into water from a bridge 44.1 m
20 m above the water. It hits the water with a certain above the water. Another stone is thrown vertically down-
velocity and then sinks to the bottom of the lake with the ward 1 s later. Both strike the water simultaneously.
same velocity 6 sec after it is dropped. [g = 10 m/s2]. What was the initial speed of the second stone ?
Find the depth of the lake. 2´ 44.1
Ans : 80 m Sol: t = s = 9 s = 3s,
9.8
Sol : Velocity on reaching water  2  10  20  20 ms 1 1
44.1 = v´ 2 + ´ 9.8´ 2´ 2
2
2h 2  20 or 2v = 44.1 - 4.9 ´ 4 = 24.5
Time taken to reach water =   2sec
g 10
24.5 - 1
\ depth of water (s) = vt = 20 (6-2) = 80 m or v = m s = 12.25 ms -1
2
Problem : 4. 81 Problem : 4.85
A particle is dropped from point A at a certain height A ball is dropped from the top of a building. It takes
from ground . It falls freely and passes through three 0.5s to fall past the 3m length of a window some distance
points B,Cand D with BC=CD . The time taken by the from the top of the building. If the velocity of the ball at
particle to move from B to C is 2 seconds and from C to D the top and at the bottom of the window are VT and VB
respectively thenVT +VB = ?
is 1 second . Find the time taken to move from Ato B ?
æu + v ö÷
Sol : Let AB=y:BC=CD=h and tAB=t Sol. S = çççè ÷t
A 2 ø÷ V
1 2 y T
then y = gt B æv + vB ö÷
2 3 = çç T ÷0.5
3m
h çè 2 ø÷ VB
1 2 C
y+ h= g (t + 2)
2 h VT + VB = 12 m / s
D
1 2 Problem : 4.86
and y + 2h = g (t + 3)
2 Two balls are projected vertically upwards with
solving these three equations , we get t=0.5 s velocitees u1 and u2 from the ground with a time gap of
n seconds. Find the time after which they meet
Problem : 4. 82
Sol. If they meet at a height h then
A ball is thrown vertically upward with a velocity 'u' 1 1
h = u1t - gt 2 = u 2 (t - n )- g (t - n )2
from the balloon descending with velocity v. After what 2 2
time, the ball will pass by the balloon ? 1 1 2 1 2
u1 t - gt 2 = u 2 t - u 2 n - g t - gn + gtn
1 2 2 2 2
Sol : Sr = ur t + ar t
2 1
u 2 n + gn 2
t= 2
1 2
O = (v + u )- gt D u + gn
2
2 (v + u ) u n
t= b) t = g + 2 if u1 = u2
g

Problem : 4.87 iii) A javelin thrown by an athlete

A balloon starts from rest from the ground and moves iv) A jet of water from a rubber tube impelled into
with uniform acceleration g/8. When it reaches a height
h a ball is dropped from it the time taken by the ball to air
reach the ground is Note 4.35 : In two dimensional motion, the position
2 gh gh of a particle is represented by the position vector
Sol : v = = v 
8 2 r  xi  yj
1 2 d dx dy
- h = vt - gt h h \ velocity = V = (xi + yj)= i + = Vxi + Vy j
2 dt dt dtj
gh 1
- h= t - gt 2 u=0 Vx and Vy are components of velocity and
2 2
represent speeds along x and y directions. These two
1 2 gh component speeds are independent of each other.
gt - t- h= 0
2 2
Simplifing and taking only the positive value as The acceleration vector is given by
negative value of t is not acceptable we get
dv d dv dv y
h a  V x i  V y j   x i  j  axi  a y j
t= 2 dt dt dt dt
g
Again the two components of acceleration are
Problem : 4.88 independent of each other
A boy sees a ball go up and then down through a window 4.30 THE TRAJECTORY OF
2.45m high. If the total time that ball is in sight in 1s,
the height above the window the ball rises is PROJECTILE IS A PARABOLA
approximately (in absence of air resistance)
Sol : Time during upward crossing of 2.45m
Let a body be projected at ‘O’ with an initial
1
= time during downwad crossing = s velo city u t hat makes an angle  with t he
2
1 2 u2 X–axis.
h = ut + gt H=
2 2g This velocity can be written as
u 
1 1 1 u = u x i + u y j = (u cos q)i + (u sin q) j .
2.45 = u. + ´ 9.8´ 1
2 2 4 h t=
2
s
u = 4.9 – 2.45 = 2.45 g = 9.8m / s
Due to the fact that two dimensional motion can
be treated as two independent rectilinear motions,
u 2 2.45´ 2.45
H= = the projectile motion can be broken up into two
2g 2´ 9.8
separate straight line motions.
H 0.3m
i) horizontal motion with zero acceleration.[i.e.,
4.29 PROJECTILE constant velocity as there is no force in horizontal
direction]
Any body projected into the air at an angle
other than 900 with the horizontal near the ii) vertical motion with constant downward
surface of the earth, is called a projectile. acceleration = g ( it is moving under gravity)
Y
The science of projectile motion is called ballistics

v x  u cos 
Examples for projectiles : x P A
u
i) A cricket ball thrown by a fielder uy
y
h

X
ii) A bullet fired from a gun O
ux C B

Let the body reach a point ‘P (x, y)’ in its For a projectile the time to reach maximum height
trajectory after time ‘t’ is called time of ascent.
i.e, horizontal displacement and vertical For a projectile, the vertical component of
displacements of the body in time ‘t’ are x and y velocity vy is zero at the highest point.
respectively. vy = u sin  –gt,
1) Let us first consider the horizontal motion. u sin 
Here, vy = 0 and t = ta  ta 
As the horizontal motion has no acceleration the g
horizontal component of projectile’s velocity u x ii) Time of flight (T) :
remains constant through out the motion, the For a projectile, the total time to reach the same
displacement of the projectile after any time ‘t’ from horizontal plane of projection is called the time of
the initial position (the origin in our case) is given by flight. It is the total time for which the projectile
x  uxt   u cos   t ............... (1) remains in air.
2) Now let us consider the vertical motion. 1 2
y   u sin   t  gt
In vertical direction, the acceleration of the 2
projectile is equal to the free fall acceleration which Here, t = T, y = 0
is constant and always directed downward 1 2u sin 
\ 0   u sin   T  gT 2 T 
a = –gj i.e., ay = – g. 2 g
The equation for vertical displacement of the Note 4.36 :
projectile after time t can be written by Time of descent = time of ascent
1 time of flight
y  u y t  a y t 2 , we get =
2 2
1
y   u sin   t  gt 2  a y   g  ..........(2) iii) Maximum height (Hmax) :
2
The vertical displacement of a projectile during
by substituting the value of ‘ t ’ from (1) as
time of ascent is the maximum height of the projectile.
x
t in equation (2) 1
u cos  In the equation y   u sin   t  gt 2 we use
2
2
 x  1  x  y = Hmax and t = ta.
y  u sin    g 
 u cos   2  u cos  
u 2 sin 2 
g \ H max 
  2 2g
 y   tan   x   2 2 x
 2u cos  
u2
Note 4.37 : When   900 , H max  .
The values of g,  and u are constants. The 2g
above equation is in the form
This is equal to the maximum height reached by
g a body projected vertically upwards.
y = ax – bx2 where a = tan  ; b  2 .
2u cos 2 
iv) Horizontal Range (R) :
This is the equation of a parabola so the path of the
projectile is a parabola. This is defined as the horizontal distance covered
by projectile during its time of flight.
4.31 MOTION PARAMETERS OF A
PROJECTILE Thus, by definition,
I) TIME OF ASCENT (ta) Range R = horizontal velocity X time of flight

2u sin  Then the magnitude of resultant velocity after time

i.e., R   u cos   T   u cos  
g
t is V  Vx 2  Vy 2 = (u cos q)2 + (u sin q - gt )2
u 2 sin 2
 Range, R  
g The velocity vector v makes an angle a with
the horizontal given by
 Vy 
a  tan 1   at this instant.
 Vx 
H
y

Range
Note 4.38 : For a given value of projection velocity a vx

u, R is maximum when vy v
0 0 x
2  90 i.e.,   45 .
So maximum horizontal range is At any vertical displacement ‘h’, velocity is

Rmax 
u2
g
( sin 90 = 1) 0 V  u cosθiˆ   u 2 sin 2 θ  2 gh ˆj 
Note 4.39: For a given speed of projection ‘u’, the Note 4.40 : The horizontal component of v elocity
ranges are equal for angles remains constant all along . (since acceleration due
to gravity has no component along the horizontal.
(a) θ and  90  θ
Note 4.41: The vertical component of velocity,
(b)  45  α  and  45  α  a) Goes on decreasing during the ascent
y b) goes on increasing during the descent
c) becomes zero at the highest point
(90-) or  Note 4.42:Thus velocity of a projectile is maximum
at projection point (equal to u)
= 450
Note 4.43:velocity of the projectile is minimum at
 or (90 - )
x highest point (equal to u cos q )
v) Velocity of the Projectile at any Instant ‘t’ : Note 4.44:change in the velocity of the projectile is
The horizontal component of the projectile equal to 2u sin q
æD v y = v f - vi = u sin q - (- u sin q)ö÷
remains constant all the time. (because acceleration çç ÷
çç= 2u sin q and D v = 0 ÷
due to gravity has no component along the è x ø÷
horizontal.) Note 4.45:
 Horizontal component of velocity after any similarly change in momentum = 2mu sin q
time t is vx = ux = ucos 
Note 4.46 : average velocity of the projectile during
Vertical component of velocity after any time t is the entire journey =
vy = uy - gt = u sin q - gt. total displacement range
 =

V  Vxiˆ  V y ˆj  total time time of flight
 (u cos q )´ T
V  u cos θiˆ   u sin θ  gt  ˆj = = u cos q
T
Note 4.47 : angle between velocity and accelaration (1) u 2 sin 2 q g

of a projectile, Þ ´ 2 2
(2) 2g u sin q
a) is between 900 to 1800 during the ascent i.e
the dot product of velocity and acceleration is -ve u 2 sin 2 q g
= ´ 2
during the ascent 2g u 2sin q cos q
b) is between 00 to 900 during the descent i.e
the dot product of velocity and acceleration is +ve Tanq H 4H
Þ = Þ Tanq =
during the descent 4 R R
c) is 900 at the highest point i.e the dot prod- Note 4.52: The time of flight (T), Range (R), and
uct of velocity and acceleration is 0 at the highest angle of projection (q) are related as,
point
(gT 2 = 2 R Tanq )
Note 4.48: At the projection point ,
2u sin q
1
total energy = Eto = mu 2 (i.e, it is purely kinetic )
Sol : T = .........[1]
g
2
u 2 sin 2q
Note 4.49 :At the highest point of the projectile , and R = ........[2]
g
1 2 2 2 2
a) kinetic energy E k = mu cos q = Eto cos q (1) T 2 4u 2 sin 2 q g
2 Þ = ´ 2
2
(2) R g u sin 2q
1
2 2 2
b) potential energy E p = mu sin q = Eto sin q
2 4u 2 sin 2 q g
= 2
´ 2
c) ratio of potential to kinetic energies g u (2sin q cos q)
Ep
= = tan 2 q 2Tanq T 2
Ek Þ =
g R
Note 4.50: If 'T' is the time of flight of a projectile, Þ gT 2 = 2 R Tanq
1 2
maximum height H = gT
8 1 2
2u sin q gT Note 4.53: R tan q = 4 H = gT
Sol : We know that T = Þ u= 2
g 2sin q
2
Note 4.54: range of a projectle is maximum when
u 2 sin 2 q æ gT ö÷ sin 2 q
çç u 2 sin 2q
\ H= = ÷ ´ 2g angle of projection = 450 ( R =
2g çè 2sin q ø÷ g
, R is
g 2T 2 sin 2 q 1 2 maximum if sin 2q is maximum i.e if

= ´ = gT 2q = 900 or q = 450
4sin 2 q 2g 8
Note 4.55: range of a projectile = maximum height
Note 4.51: For a projectile, angle of projection [ q ],
range [R] and maximum height [H] are related as if q = tan- 1 4 or 760
4H
æ ö Sol : we know that tan q =
ççTanq = 4H ÷÷ R
from this
çè R ø÷
u 2 sin 2 q R = H Þ tan q = 4 Þ q = tan- 1 4 = 760
Sol : Maximum height H = .....(1)
2g Note 4.56: For projectile, in the case of complimen-
2
u sin 2q tary angles,
Range R = .....(2)
g a) Ranges are same

Sol : If  and 90    are angles of projection, we Proof: Applying v2 - u2 = 2as for upward journey
have of a projectile,
u 2 sin 2 u 2 sin 290    we have, u = u sin q , a= -g,
R1  and R2 
g g
2
u sin 2 u sin 180  2 
2 1 1 u 2 sin 2 q
 R1  and R2  S= H=
g g 2 2 2g
 R1  R2 Substituting these values we get
u2
b) If H1, H2 are maximum heights, H1 + H2 = 2 1 u 2 sin 2 q
2g v 2 - (u sin q ) = - 2 g ´ ´
Sol: 2 2g
u 2 sin 2  u 2 sin 2  90    u 2 sin 2 q u 2 sin 2 q
we have 1 H  , 2H  \ v 2 = u 2 sin 2 q - =
2g 2g 2 2
2 2
u sin  u cos  2 2 u sin q
H1  H 2   Þ v=
2g 2g 2
2
u Note 4.59:velocity of a projectile at half of maxi-
 H1  H 2 
2g 1 + cos 2 q
mum height = u
c) R1 = R2 = R = 4 H1H 2 2
2R Proof :Velocity at any instant is
d) If T1, T2 are times of flight, T1 T2 =
g v= v x2 + v y2
Sol: We have
2u sin  2u sin  90    But vx = ux = u cos q at any point
T1 = and T2 
g g u sin q
while v y = (at half of maximum height)
4u 2 sin  cos  2
 TT
1 2  æu sin q ö÷2
g2
2
\ v= (u cos q) + ççç ÷
è 2 ø÷
2
2u sin 2 Simplifying we get

g2
1 1 + cos 2 q
\ R = gT1T2 v= u
2R 2
T1T2  i 2
g Note 4.60 : The physical quantities which remains
constant during projectile motion are
Note 4.57: If horizontal and vertical displacements
of a projectile are respectively x = at and y=bt-ct2 , i) acceleration due to gravity g
1 2
then velocity of projection u = a 2 + b 2 and angle of ii) total energy E0 = mu
2
projection iii) horizontal component of the velocity
b u cos q
q = tan - 1
a Note 4.61:The physical quantities which change
Note 4.58 : For a projectile, during projectile motion are
'y' component of velocity at half of maximum i) speed ii) velocity
u sin q iii) linear momentum iv) KE
height = v) PE
2
Note 4.62:A particle is projected up from a point at 2) x-t graph is a straight line passing through
an angle with the horizontal . At any time ‘t’ if origin( x = u cos qt )
P=linear momentum y=vertical displacement 3) y-t graph is a parabola.
x=horizontal displacement , then the kinetic energy
1 2
(K) of the particle plotted against these parameters ( y = u sin q - gt )
can be : 2
4) vx-t graph is a straight line parallel to time-
k – y graph From conservation of mechanical axis ( vx = u cos q )
energy K=Kt-mgy....................(1)
Note 4.64: If air resistance is taken into consider-
(here Kc=initial kinetic energy=constant) ation then
i.e K-y graph is straight line . a) trajectory departs from parabola
K
It first decreases b) time of filght may increase or decrease
linearly becomes minimum K c) the velocity with which the body strikes the
at highest point and then ground decreases
Y d) maximum height decreases
becomes equal to Kc in the
similar manner . There e) striking angle increases
fore K-y graph will be as shown in figure f) range decreases
k- t graph : Note 4.65: A projectile is fired with a speed u at an

K
angle  with the horizontal. Its speed when its
Equation (1) can be direction of motion makes an angle  with the hori-
æ 1 2ö
written as K = K i - mg çççèu y t - gt ø÷
÷
÷
zontal
2 v = ucos  sec. 
i.e K-t graph is a parabola . Explanation : Horizontal component of velocity
t
Kinetic energy first decreases and then increases remains constant
k - x graph K  v cos   u cos  v = ucos  sec. 
Equation (1) can also written as Note 4. 66 : A body is dropped from a tower. If
æ wind exerts a constant horizontal force the path of
gx 2 ö÷ x
K = Ki - mg çççx tan q - ÷
2÷ K the body is a straight line
çè 2ux ø÷
again K-x graph is a parabola Note 4.67 : The path of projectile as seen from
another projectile:
k - p2 graph 1
x1 = u1 cos q1t y1 = u1 sin q1t - gt 2
p2 2
2
Further p =2Km 2
i.e p a K 1
x2 = u2 cos q2t y2 = u2 sin q t - gt 2
2
or K versus p2 graph is a straight line passing through
D x = (u1 cos q1 - u2 cos q2 )t
origin
D y = (u1 cos q1 - u2 cos q2 )t
Note 4.63: In a projectile motion let vx and vy are the
horizontal and vertical components of velocity at any D y u1 sin q1 - u2 sin q2
=
time t and x and y are displacements along horizontal D x u1 cos q1 - u2 cos q2
and vertical from the point of projection at any time
t . Then i) If u1 sin q1 = u2 sin q2
1) vy-t graph is a straight line with negatiave i.e., initial vertical velocities are equal slope
Dy
slope and positive intercept = 0
Dx
( v y = u sin q - gt ) Þ the path is horizontal straight line
ii) If u1 cos q1 = u2 cos q2 a parabolic path. The time taken to reach ground is
i.e., initial horizontal velocities are equal arrived as explained below.
Dy u cos q
q
slope =¥ u
Dx u sin q
Þ the path is a vertical straight line

iii) u1 sin q1 > u2 sin q2
u1 cos q1 > u2 cos q2 The components of velocity are as shown. Here
Þ the path is a straight line with + ve slope the body can be treated as a body projected vertically
down with a velocity u sin q from a tower of height
iv) u1 sin q1 > u2 sin q2 ; u1 cos q1 < u2 cos q2
h. Hence the equation of motion on reaching the foot
or u1 sin q1 < u2 sin q; u2 cos q1 < u2 cos q2 of the tower is
Þ the path is straight line with –ve slope 1 2
h  (u sin  )t  gt
Note 4.68 : Two bodies thrown with same speed 2
from the same point at the same instants but at 1 2
different angles can rever collide in air (using S  u1t  gt where u1  u sin  )
2
1 2
 (x = u cos q t , , y = u sin q - gt , x, y Problem : 4.89
2
A stone is to be thrown so as to cover a horizontal
coordinates always differ)
distance of 3 m. If the velocity of the projectile is 7
Note 4.69 :A body is projected up with a velocity u ms–1, find
at an angle q to the horizontal from a tower of height (i) the angle at which it must be thrown,
h as shown. It is clear that such a body also traces a (ii) the largest horizontal displacement that is possible
parabolic path. The time taken to reach ground is with the projection speed of 7 ms-1.
arrived as explained below . Sol. (a) Given that, u=7ms–1, R = 3m
u sin q
u 2 sin 2 (7)2 sin 2
R 3
u g 9.8
q
u cosq  sin 2 θ = 3/5
h  2 θ = 37o or 180o – 2 θ
 θ = 18o 30' or θ = 71o 30'
Hence a range of 3m is possible with two angles of
The components of velocity are as shown. projections.
Here the body can be treated as a body projected (b) For maximum range with a given velocity, the angle
vertically up with a velocity u sin q from a tower of of projection, q = 45o
height h . Hence the equation of motion on reaching (7)2 sin 2(450 )
R max   5m
the foot of the tower is 9.8
Problem : 4.90
1 2
h  (u sin  )t gt The speed with which a bullet can be fired is 150 ms-1.
2
Calculate the greatest distance to which it can be
by using the formula for height of tower projected and also the maximum height to which it
Note 4.70 : A body is projected down with a velocity would rise.
u at an angle q to the horizontal from a tower of u 2 sin 2a u 2 sin 2 a
Hint : R = , H max = Here, a = 450
height h as shown. It is clear that such a body traces g 2g
Ans. 2295.14m, 573.97m

A cannon and a target are 5.10 km apart and located If at point of projection, the velocity of a particle is
at the same level. How soon will the shell launched "u" and is directed at an angle "  " to the horizontal,
with the initial velocity 240 m/s reach the target in the then show that it will be moving at right angles to its
absence of air drag? (u cosec  )
initial direction after a time
Sol. Here, v0 = 240 ms–1, R = 5.10 km = 5100 m, g
g = 9.8 ms–2,  = ? Sol. Let "t" be the time after which velocity becomes
perpendicular to its initial direction.
v 20 sin 2 As u and v are perpendicular, the angle between v and
R=
g vertical will be a.
Rg p
sin 2 α = u
2
v0 
v
 α = 30° or 60° O

2v0 sin 
using = T =
g Initial velocity 'u'  (u cos  î  u sin  ˆj)
When, α = 30°, T1 =
2  240  0.5
= 24.5 s

After t sec, velocity 'v'  u cos  î  (u sin   gt) ˆj 
9.8  These are perpendicular their dot product is zero.
2  240  0.867  (u cos  i  u sin j)   u cos  i  (u sin   gt) j  0
When, α = 60°, T2 = = 42.41 s
9.8
u cos ec 
Problem : 4.92 and t 
g
The horizontal range of a projectile is 2 3 times its Problem : 4.95
maximum height. Find the angle of projection. Find the velocity in the above problem at the instant
when the instantaneous velocity is perpendicular to
Sol. Hint R tan q = 4 H ;
velocity of projection
æ2 ö Sol. From the figure its clear that angle made by the
Ans : q = tan ççç ÷
- 1
÷
è 3 ø÷ instantaneous velocity vector with horizontal is
Problem : 4.93 900 - a
p
The ceiling of a long hall is 20 m high. What is the u
maximum horizontal distance that a ball thrown with 
v
a speed of 40 ms–1 can go without hitting the ceiling of 
O
the hall (g = 10 ms–2) ?
Sol. Here, H = 20 m, u = 40 ms–1. since the horizontal component of velocity does not
Suppose the ball is thrown at an angle  with the change, we have,
horizontal. ( )
v cos 900 - a = u cos a Þ v sin a = u cos a
u 2 sin 2  (40)2 sin 2  Þ v = u cot a

Now, H = Þ 20 = Problem : 4.96
2g 2 10
or, sin  = 0.5 or,    A particle is projected from the origin in X–Y plane.
Acceleration of particle in Y direction is  . If equation
u 2 sin 2 (40)2  sin 120° of path of the particle is y = ax – bx2, then find initial
Now R = =
g 10 velocity of the particle.
Sol. y = ax – bx2
(40) 2  0.866 x2
  138.56cm
10 y = x tan  – 2
2u cos2 
 u 2 sin 2q 2u sin q
tan  = a and 2 2
b ii) R = iii) T =
2u cos  g g
Ans : i) 2.1m ii) 10m iii) 1.414s
u=

 1 a2 .
2b Problem : 4.100
A body is projected with velocity u at an angle of
Problem : 4.97
projection  with the horizontal. The body makes 300
A particle is projected from the origin. If y = ax – bx2, with the horizontal at t = 2 second and then after
is the equation of the trajectory then find 1 second it reaches the maximum height. Then find
i) angle of projection ii) range (a) angle of projection (b) speed of projection
iii) maximum height Sol : During the projectile motion, angle at any instant t is
Sol. comparing the given equation with the equation of the suchthat
trajectory of a projectile given by usin   gt
gx 2 tan  =
ucos 
y = x tan q - ,
2u 2 cos 2 q For t = 2seconds, a = 30 0
-1
i) we have tan q = a Þ angle of projection q = tan a
1 u sin   2 g
ii) For y = 0 to the x coordinat gives range   (1)
3 u cos 
a
\ R=
b For t = 3 seconds, at the highest point a = 0 0
4Η \ a2 u sin q - 3 g
iii) tanθ = H= \ 0=
R 4b u cos q
Problem : 4.98
usin 
3 
1 2 g or u sin  = 3g _______(2)
If y = x – x is the equation of a trajectory, find the
2
using eq. (1) and eq.(2)
time of flight.
1 u cos  = 3g _______(3)
Sol. We have y = x – x2 = x (1 – x/2)
2
If y = 0, then either x = 0 or x = 2. Eq.(2)  eq.(3) give q = 60 0 squaring and adding
equation (2) and (3)
Hence the range of the motion is 2.
u = 20 3 m/s.
For half the range, x = 1, then y = 1/2
Hence maximum height attained H = 1/2. Problem : 4.101
Time to reach maximum height, A ball is thrown from the top of a tower of 61 m high
with a velocity 24.4 ms–1 at an elevation of 30° above
2H 1
ta =  the horizontal. What is the distance from the foot of the
g g
tower to the point where the ball hits the ground ?
1
Time of flight, T = 2t = 2 Sol : (Hint) :
g
u sin q
*Problem : 4.99
u
A ball is thrown with velocity 10ms–1 at an angle q
u cos q
of   45 to the horizontal. Find i) the height of which
h
the ball will rise to ii) the distance x from the point of
projection to the point where it reaches to the ground
and iii) the time during with the ball will be in motion
(neglect the air resistance) (g = 10ms–2) 1 2
h= gt - (u sin q)t
u 2 sin 2 q 2
Þ t = 5sec onds
Hint : i) H max =
2g Also, d = (u cos q)t = 105.65 m

Problem : 4.102 Thus, in time t the bullet passes through A a vertical

1 2
A particle is projected at an angle of elevation  and distance gt below M.
2
after t secons it appears to have an elavation  as seen The vertical distance through which the monkey fall in
from the point of projection. Find the initial velocity 1 2
of projection. time t. s  gt
1 2 2
y u sin  t  2 gt Thus, the bullet and the monkey will always reach at
Sol. tan    point A at the same time.
x u cos  t
gt
u(sin   cos  tan )  Problem : 4.105
2
gt A particle is projected from the ground with an initial
u
2(sin   cos  tan ) speed u at an angle  with horizontal. What is the
average velocity of the particle between its point of
gt cos  projection and highest point of trajectory ?
u
2 sin(   )
Total displacement
Sol. Vav 
Problem : 4.103 Total time
y
A rifle with a muzzle velocity of 100 ms-1 shots a bullet
2
at a small target 30 m away in the same horizontal R R/2, H
 H2
line. How much height above the target must the rifle Vav  2
usin  
be aimed so that the bullet will hit the target?
g
u 2 sin 2q 2
Hint : R = Ans. q = 0.015  u2   u2 2 
g  sin2   4  sin  
R2  4H2  g   2g 
Problem : 4.104 Vav  
2usin  2usin 
A hunter aims his gun and fires a bullet directly at a g g
monkey on a tree. At the instant the bullet leaves the
barrel of the gun, the monkey drops. Will the bullet hit u2 4sin 2  cos 2   sin 4  u
  1  3cos 2 
the monkey ? g 2u sin  2
Sol. Horizontal distance travelled g
x Problem : 4.106
OB = x = u cos q t or t 
u cos  A particle is thrown over a triangle from one end of a
For motion of bullet from O to B, the vertical
horizontal base and grazing the vertex falls on the
1 2 other end of the base. If  and  be the base angles
height AB  u sin  t  gt
2 and  be the angle of projection, prove that
 x  1 2
 u sin    gt tan  = tan  + tan  .
 u cos   2
gt 2 Sol : The situation is shown in figure. From figure,we have
 x tan   ......(i)
2
Also from figure MB  x tan  Y R=Range
Now the height through which monkey falls A(x, y)
y  MA  MB  AB
 gt 2   1 gt 2  h
 x tan    x tan     
 2  2 M x
X
O R-x
y y
tan   tan  
A x Rx
u
yR
O  tan   tan  
ucos B
x
x R  x  –––––– (1)
Gun x

 x Sol. Let  be the angle of projection and u be the velocity of

But equation of trajectory is y  x tan  1
 R  projection. It is given that the maximum height of the
yR projectile is 2 h, we have
tan  
x R  x  usin   4gh
From Eqs. (i) and (ii), tan   tan   tan  If time taken by the stone to reach points A and B are t1
Problem : 4.107 and t2, then t1 and t2 are the roots of the equation
Two shots are fired simultaneously from the top and 1 2
h  u sin  t  gt2 or gt  2ut sin   2h  0
bottom of a vertical tower AB at angles  and  with 2
horizontal respectively. Both shots strikes at the same
point C on the ground at distance ‘S’ from the foot of usin  u2 sin2   2gh
Solving, t  
the tower at the same time. Show that the hight of the g g
tower is S  tan   tan .
U1 Using u sin   4 gh

A
4h 2h
t 
h
U2 g g
4h 2h
 C Thus, we have t1  
B S g g
Sol. S   u1 cos   t   u2 cos   t 4h 2h

and t 2  
S S g g
t  
u1 cos  u2 cos  Now, the distance AB can be written as
Let, height of tower be h vt2  ucos   t2  t1 
1 1 (v = velocity of the bird)
 h   u1 sin   t  gt 2 0   u2 sin   t  gt 2
2 2
Ratio of horizontal velocities
or  u1 sin   t  h   u2 sin   t
v t t 2
S u sin   2 1
 u1 sin . h 2 .S ucos  t2 2 1
u1 cos  u2 cos 
Problem : 4.109
h  S tan   S tan 
The velocity of a projectile when at its greatest height
 h  S  tan   tan  
2
is of its velocity when at half of its greatest height
Problem : 4.108 5
find the angle of projection
A stone is projected from the point of a ground in such
a direction so as to hit a bird on the top of a telegraph Sol : Step 1 : we know that, velocity of a projectile at half of
post of height h and then attain the maximum height 1 + cos 2 q
2h above the ground. If at the instant of projection, the maximum height = u
2
bird were to fly away horizontally with a uniform speed,
find the ratio between the horizontal velocities of the 2 1 + cos 2 q
bird and the stone, if the stone still hits the bird while Step 2 : given that u cos q = ´ u
5 2
descending.
Squaring on both sides
Y
2 æ1 + cos 2 q ö÷
u 2 cos2 q = u 2 çç ÷
A B
5 çè 2 ø÷
÷
u
10 cos 2 q = 2 + 2 cos 2 q
h 2h 1
 Þ 8 cos 2 q = 2 Þ cos 2 q = Þ q = 600
X 4
O


Afoot ball is kicked of with an initial speed of 19.6 m/ A body projected from a point ‘O’ at an angle q , just
sec at a projection angle 450. A receiver on the goal crosses a wall ‘y’ m high at a distance ‘x’ m from the
line 67.4 m away in the direction of the kick starts point of projection and strikes the ground at O’ beyond
running to meet the ball at that instant. What must his the wall as shown, then find height of the wall ?
speed be if he is to catch the ball before it hits the
ground ? Y R=Range
2
u 2 sin 2q (19.6) ´ sin 90
Sol : R = =
g 9.8
 y
or R = 39.2 metre. X
O x R-x O'
Man must run 67.4 m - 39.2 m = 28.2 m
in the time taken by the ball to come to ground.  x
Time taken by the ball. Ans : y  x tan  1 
 R
2u sin q 2´ 19.6´ sin 450 4 Sol. We know that the equation of the trajectory is
t= = =
g 9.8 2 gx 2
y = x tan q - can be written as
t = 2 2 = 2 ´ 1.41 = 2.82sec. 2u cos 2 q
2
28.2m æ gx 2 ö÷sin q
Velocity of man =
2.82sec
= 10 m / sec. y = x tan q - çç 2 ÷
÷sin q
çè2u cos 2 q ø÷
Problem : 4.111
gx 2 tan q
A projectile has the maximum range of 500m. If the y = x tan q - 2
projectile is now thrown up on an inclined plane of 300 u (2sin q cos q )
with the same speed, what is the distance covered by it x 2 tan q
along the inclined plane ? Þ y = x tan q - 2
u sin 2q / g
u2
Sol : Rmax =
g  x u 2 sin 2q
Þ y  x tan  1 R  [since R = ]
u2   g
 500  or u  500 g
g Problem : 4.114
v 2 - u 2 = 2 gs A body is projected with a velocity 'u' at  to the
0 - 500 g = 2 x (-g sin 300 ) x x horizontal . Find radius of curvature of the trajectory
x = 500 m. when the velocity vector makes  with horizontal.
Problem : 4.112
Sol: Let v be the velocity of particle when it makes  with
horizontal . Then
Two stones are projected with the same speed but mak-
u cos q
ing different angles with the horizontal. Their ranges v cos a = u cos q or v =
are equal. If the angle of projection of one is  / 3 and cos a
Y
its maximum height is y1, then what is the maximum
height of the other ?
Sol : Here speeds of projection and ranges are same and hence v
u
angles of projections are 
    u cos 
and   
3 2 3 6 
X
y 2 u 2 sin 2  2 2g g g cos 
Now   2
y1 2g u sin 2  1
sin 2  2 sin 2  / 6  1 y1 it is clear that g cos a plays the role of radial accel-
   y2 
sin 2  1 sin 2  / 3 3 3 eration

v2 v2 Problem : 4.118
g cos a = Þ R=
R g cos a A projective of 2kg was velocities 3m/s and 4m/s at
2 two points during its flight in the uniform gravitational
æu cos q ö÷ æçç 1 ÷=
ö÷ u 2 cos 2 q
= çç ÷ field of the earth. If these two velocities are ^ to each
çè cos a ø÷ çè g cos a ø÷
÷ g cos 2 a
other then the minimum KE of the particle during its
Note : When the projectile is at the highest point, its clear flight is
that  =00 . V1 cos a = V2 cos (90 - a )
3
2 2 3cos a = 4 sin a
u cos 
R= 3 a 900 - a
g tan a =
4
Problem : 4.115 4
1 2
For a projectile , projected with a velocity u at an KEmin = mv1 cos 2 a
2
angle q to the horizontal . Find the magnitude of torque
about the origin when it strikes the ground 1 æ4 ö2
= ´ 2 ´ 3´ çç ÷ ÷
Sol. we know that torque 2 èç 5 ÷
ø
t =force x perpendicular distance from the origin on to 9´ 16

the line of action of force = = 5.76 J
25
  
mg ´ u 2 sin 2q Note 4.71 : If u = xi + yj
= mgxrange =
g  
i along horizontal j along vertical
y2 2y 2 xy
H= , T= , R=
q 2g g g
Range (R) q
   
mg
Note 4.72 : u = ai + b j + ck
  
Problem : 4.116 i –east j - north k -vertical
A grass hopper can jump maximum distance of 1.6m.
It spends negligible time on the ground. How far can ux = a 2 + b 2 uy = c
it go in 10 seconds?
2c c2
u2 T= ;H = ,
= 1.6 u 2 = 16 u = 4m / s g 2g
g
1
4 cos q = 4´
2
= 2 2m / s
R=
2 ( a 2 + b2 c )
g
S = 4 cos q.t = 2 2 ´ 10
Problem : 4.119
S = 20 2m
Wind imparts a horizontal acceleration of 0.4m/s2
Problem : 4.117 towards left. q = ? for the ball to fall in the hand of
A particle is projected with a velocity of 10 2 m/s at thrower
an angle of 450 with the horizontal. Find the interval
2u 2 sin q cos q
between the moments when speed is 125 m / s . Sol. R = = u xT
g
Sol. (g = 10m/s2) S 2u cos q
un = 10, uy = 0 q T=
g
v 2 = vx2 + v 2y
Dt 1 2
125 = 100 + v 2y R | = u xT - aT
2
vy = 5
2v 2 ´ 5 2u 2 sin q cos q 1
Dt = = = 1s O= - 0.4.T 2
g 10 g 2

2
2u 2 sin q cos q 1 4 æ2´ 2u cos q ö÷ 500 4
= ´ ´ çç ÷ = ´ ´ 10
g 2 10 çè g ø÷
÷ 3 5
- 1 4000
Tan q = 0.4  q = tan (0.4) = m
3
In the absence of wind the range and maximum height A golfer standing on the ground hits a ball with a velocity
of a projectile were R and H. If wind imparts a
5
horizontal acceleration a = g/4 to the projectile then of 52 m/s at an angle q above the horizontal if tan q =
find the maximum range and maximum height . 12
find the time for which the ball is atleast 15m above the
Sol : H 1 = H (u sin q remains same)
T| = T (
ground? g = 10 m
s2 )
1
R| = ux T + aT 2 Sol. v y = 2
u y - 2gy 15m
2
1 g 2 5´ 5
= R+ T = 52´ 52´ - 2´ 10 ´ 15
2 4 13´ 13
1
= R+ gT 2 = 16´ 25 - 300 = 10
8
2u y 2´ 10
= R+ H Dt = = = 2s
10 10
R1 = R + H H1 = H
4.32 HORIZONTAL PROJECTION
Problem : 4.121 FROM THE TOP OF A TOWER :
  
u = 4i + 4 j . mass = 2kg. A constant force Equation for Path (Trajectory) :
F = - 20 jN acts an the body. Initially the body was Suppose a body is projected horizontally with
at (0,0). Find the x coordinate of the point where its y
an initial velocity u from the top of a tower of height
coordinate is again zero.
‘h’ at time t=0. As there is no horizontal acceleration,
20
Sol : a = = 10 the horizontal velocity remains constant throughout
2
the motion.
2u x u y 2´ 4´ 4 Hence after time t , the velocity in horizontal
R= =
10 10 direction will be vx = u.
R = 3.2m 0
u
Problem : 4.122 y vx= u

P
h x
A particle is projected from a tower as shown in figure,
then find the distance from the foot of the tower where it vy= gt
will stike the ground. (g = 10 m/s2)
A Horizontal projectile B
1 2
Sol. s = ut + at 370
2 Let the body reach a point ‘P’ in time t. Let x
1500 m 500
500 1 m/s and y be the coordinates of the body.
1500 = sin 37 + 10 ´ t 2 3
3 2 For the y-coordinate, after time t seconds
500 3
1500 = ´ t + 5t 2 1 2 1
3 5 y gt [ y  u y t  a y t 2 ]............. (1)
2 2
300 = 20t + t2
For x – coordinate, after t seconds
On solving , t = 10 s
x = ut ( the horizontal velocity is constant)
\ horizontal distance = u cos q. T ............ (2)
 t  x/u
From Eqs. (1) and (2) we get The velocity along Y–axis is
1 x
2
æg ö 2 v y = u y  gt and u y = 0 as the body is thrown
\ y = çç 2 ÷ ÷x .......... (3)
y g 
2 u  çè 2u ø÷ horizontally initially.
 v y  gt
g
g and u being constants,  2  is a constant. u x
 2u  O
g 2 y
If 2
 k then y = kx .
2u x
h Vx = u
This equation represents the equation of a P q
parabola. Vy V
4.33. MOTION PARAMETERS OF A R V
Y
HORIZANTAL PROJECTILE
So, the magnitude of the velocity
i) Time of Descent :
It is the time the body takes to touch the ground V= v 2 x + v2 y =u 2 + g 2t 2

after it is projected from the height ‘h’. If velocity vector v makes an angle  with the
For y = h and t = t we get vy gt
d horizontal then tan    (or)   tan 1  gt 
vx u u
1 2 2h
h= gtd \ td = Note 4.73:
2 g
For an easier understanding, we consider that,
The time of descent is independent of initial [Motion of Horizontal projectile = Motion in
velocity with which the body is projected and depends y-direction like a freely falling body + Motion in
only on the height from which it is projected. x-direction with constant velocity.]
Note : td is the time of flight in this case. Application 4.20 :
ii) Range : If a body projected horizontally with velocity

u from the top of a tower strikes the ground at
The maximum horizontal distance travelled by an angle of 450
the body while it touches the ground is called range
Vy=Vx gt=u u
(R). It is shown as AB in the Fig.

As the horizontal velocity is constant u
\ t= Vx=u
Range = R = horizontal velocity x time of g 450
Vy = 2gh = u
descent = (u)t . Application 4.21

d
A body is projected horizontally from the
2h 2h
But td   R  u  top of a tower. The line joining the point of
g g projection and the striking point make an angle
iii) Velocity of the projectile at any time ‘t’ of 450 with the ground. Then h=u
Let the body be at point P after the time t. 1 2 u
gt = ut
Let v x and v y be velocities along x and 2
y–directions. h
2u
The horizontal velocity remains constant t=
g 450
throughout the motion. Hence vx = u.
x
Application 4.22 : Application 4.24

A body is projected horizontally from the top An aeroplane flies horizontally with a
of a tower . The line joining the point of projection velocity 'u' . If a bomb is dropped by the pilot
and the striking point make an angle of 450 with when the plane is at a height 'h' then
the ground . Then, the displacement = a) the path of such as body is a vertical straight
2h or 2 X line as seen by the pilot and
b) The path is a parabola as seen by an observer
A on the ground
u
c) the body will strike the ground at a certain
horizontal distance. This distance is equal to the
h
2u
450 range given by x = ut = u
g
B x C
Application 4.25:A ball rolls off the topof a
staircase with a horizontal velocity u. If each step
0 h
From the figure tan 45  1  h X has height h and width b, then the ball will just
X
hit the nth step if n equals to
2 2
 displacement AC = (AB ) + (BC ) 1 2
Sol :  nb = ut and nh  gt
2
u
= h2 + H 2 = 2h or 2 X (sin ce h = X ) 2hu 2 h
n 1
gb 2 b
Application 4.23 : 2
3
Two towers having heights h 1 and h2 are
n
separated by a distance ‘d’. A person throws
R
a ball horizontally with a velocity u from the top
Application 4. 26 :
of the
From the top of a towerone stone is thrown
1st tower to the top of the 2nd tower, then
towards east with velocity u1 and another is thrown
Time taken,
towards north with velocity u2. The distance between
2  h1  h2 
t then after stihing the ground. d = t u12 + u22
g
Application 4.27: Two bodies are thrown horizon-
u
tally with velocities u1, u2 in mutually opposite di-
(h1-h2) rections from the same height. Then
a) time after which velocity vect ors are
h1 h2 u1 u 2
perpendicular is t = .
g
For velocity vectors to be perpendicular after a
d
time t, their dot product must be zero.
Distance between the towers \ v1.v2  0
2  h1  h2  (u iˆ -
1 gtjˆ)(
. - u2iˆ - gtjˆ)= 0
d  ut  u u1u2
g t 
g
b) Separation between them when velocity 2

 x 2   x tan   x 1  tan 2 
vectors are perpendicular is  x sec 
 u1  u2  u1u2 1 2 y 1 gt 2
X   u1  u2  t  x = ut ; y = gt ; =
g 2 x 2 ut
c) Time after which their displacement vectors gt 2u
tan q = ; t= tan q
2u g
2 u1u2
are perpendicular is t  2u 2 2u 2
g x = ut = tan q ; \ R= tan q sec q
For displacement vectors to be perpendicular then g g
their dot product must be zero Problem : 4.126
æ öæ ö
ççu1tiˆ - 1 gt 2 ˆj ÷÷. çç- u2tiˆ - 1 gt 2 ˆj ÷
÷ = 0
An aeroplane is flying in a horizontal direction with a
èç 2 ø÷ èç 2 ÷
ø velocity of 600 km/hour at a height of 1960 m. When it
is vertically above a point A on the ground, a body is
2 u1u2 dropped from it. The body strikes the ground at a point
t
g B. Calculate the distance AB.
d) Separation between them when displacement
2h
Hint : t = , R = ut
is perpendicular to X   u1  u2  t  
u1  u 2  2 u1u2 g
g Ans : 20s, 3.33 km
Two paper screens A and B are separated by a
Two particles move in a uniform gravitational field
distance of 100 m. A bullet pierces A and then B. The with an acceleration "g". At the initial moment the
hole in B is 10 cm below the hole in A. If the bullet is particles were located at same point and moved with
travelling horizontally at the time of hitting the screen velocities u1 = 0.8 ms-1 and u2 = 4.0 ms-1 horizontally in
A, calculate the velocity of the bullet when it hits the opposite directions. Find the distance between the
screen A. Neglect the resistance of paper and air. particles at the moment when their velocity vectors
Sol. The situation is shown in Fig. become mutually perpendicular. (g = 10 ms–2)
Hint : v 1P v 2
2 (h1 - h2 ) u P x Q 1 2
d= u 90
o
g 0.1m
2´ 0.1 R gt
100 = u
9.8 A
100m
B
u = 700m / s u1u2
t=
g
, x = (u1 + u2 )t
Problem : 4.125
A particle is projected horizontally with a speed "u" Ans : 0.48m
from the top of plane inclined at an angle "  " with the
Problem : 4.128
horizontal. How far from the point of projection will the
A boy aims a gun at a bird from a point, at a horizontal
particle strike the plane?
distance of 100m. If the gun can impart a velocity of
u
500 m/sec to the bullet, at what height above the bird

must he aim his gun in order to hit it?
y
R (take g = 10 m/sec2)
Sol. 
Sol : x = vt or 100 = 500 × t
x
t = 0.2 sec.
x
1 2
æy ö Now h = 0 + 2 ´ 10 ´ (0.2)
çç = tan q÷÷
R  x2  y2 çè x ø÷ = 0.20 m = 20 cm.

Problem : 4.129 x Short Answer Questions

A staircase contains three steps each 10 cm high and
20 cm wide as shown in the figure. What should be the 1. If the average velocity of an object is zero in
minimum horizontal velocity of a ball rolling off the some time interval, what can you say about the
uppermost plane so as to hit directly the edge of the displacement of the object for that interval ?
lowest plane ? [g = 10 m/s2]
2. Can the instantaneous velocity of an object
Sol : h = 30 cm a
during a time interval ever be greater in
2h b magnitude than average velocity over the entire
S  60cm  v
g interval ? Can it be less ?
2  30 3 h
v v 3. Give examples for the particles where (a) the
1000 50
velo city is in opposite direction to the
60 50 3 2
v   5  3 2 acceleration (b) the velocity of the particle is
100 3 5 3
=2.45 m/s zero but its acceleration is not zero.
Problem : 4.130 4. Can equations of Kinematics be used in a
An enemy plane is flying horizontally at an altitude of situation where the acceleration varies in time ?
2 km with a speed of 300 ms-1 .An armyman with an Can they be used when acceleration is zero ?
anti - aircraft gun on the ground sights ht enemy plane
when it is directly overhead and fires a shell with a 5. A particle moves rectilinearly with uniform
muzzle speed of 600 ms-1. At what angle with the verti- acceleration. Its velocity at time t = 0 is v1 and
cal should the gun be fired so as to hit the plane ?
at time t = t is v2. The average velocity of the
Ans : 300
v1  v2
Sol. Let G be the position of the gun and E that of the enemy particle in this time inteval is . Is this
plane flying horizontally with speed 2
statement true or false ? Substantiate your
u = 300 ms-1 , when the shell is fired with a spped v0 is
answer.
vx = v0 cos q
Let the shell hit the plane a point P and let t be the time 6. Show that time of ascent of a vertically projected
taken for the shell to hit the plane. It is clear that the body is equal to time of descent.
shell will hit the plane, if the horiazontal distance EP
travelled by the plane in time t = the distance travelled 7. In case of an object vertically projected, show
by the shell in the horizotal direction in the same time, that the speed of the object upon reaching the
i.e. point of projection is equal to the speed of
u ´ v = v x ´ t or u = v x = v0 cos q projection.
u 300 8. Show that the trajectory of an object thrown at
or cos q = =
v0 600
a certain angle with the horizontal is a parabola.
E u PH2
9. Explain the terms the average velocity and
instantaneous velocity. When are they equal?
vy
0 10. Show that the maximum height and range of a
(90 - q )
q Ground u 2 sin 2  u 2 sin 2
vx pro jectile are and
2g g
= 0.5 or q = 600 . respectively where the terms have their regular
Therefore, angle with the vertical = 900 -q = 300 . meanings.

11. Show that the trajectory of an object thrown 20. When two stones are thrown from the top of
horizontally from certain height is a parabola. tower, one vertically upwards with a speed u
and the second vertically downward with a
12. Can the velocity of an object be in a direction
speed u, show that the two stones will reach the
other than the direction of acceleration of the
ground with the same speed.
object ? Explain.
21. How the horizontal and vertical component of
13. A stone is thrown up in the air. It rises to a
velocity of a projectile vary with time during
height h and then returns to the thrower. For
the motion ?
the time that the stone is in air, sketch the
following graphs: y versus t; v versus t; a versus x Very Short Answer Questions
t.
1. Give two examples of the motion of big objects
14. The figure below shows four graphs of x versus where the object can be treated as a particle and
time, which graph shows a constant, positive, where it can not be.
non-zero velocity ?
2. The state of motion is relative. Explain.
3. How is average velocity different from
instantaneous velocity ?
4. If instantaneous velocity does not change from
15. If the above four graphs have ordinate axis instant to instant will the average velocities differ
indicating velocity v and abscissa time t which from interval to interval ?
graph shows (a) constant and po sitive 5. Can an object have (i) a constant velocity even
acceleration, (b) constant and negative though its speed is changing ?
acceleration, (c) a changing acceleration that is (ii) a constant speed even though its velocity is
always positive and (d) a constant velocity? changing ?
16. Show that for a projectile launched at an angle 6. Give an example of a case where the velocity
of 450 the maximum height of the projectile is of an object is zero but its acceleration is not
one quarter of the range. zero.
17. A bird holds a walnut between its bills takes it 7. Give an example of a motion for which both
high above ground. While flying parallel to the acceleration and velocity are negative
the ground it lets the nut go off. (a) What is the
8. ‘Speed of a particle can be negative’–Is this
trajectory of the nut with respect to the bird and
statement correct ? If not why ?
(b) as seen by an observer on the ground?
9. What is the acceleration of a projectile at the
1 2 top of its trajectory ?
18. Derive the equation s  ut  at using
2
10. Can a body in free fall be in equilibrium ?
graphical method where the terms have their
usual meaning. Explain.
19. Represent graphically the motion of a body 11. If the trajectory of a body is parabolic in one
starting from rest and moving with uniform reference frame, can it be parabolic in another
acceleration both in terms of velocity-time and reference frame that moves at constant velocity
displacement-time. with respect to the first reference frame ? If the
trajectory can be other than parabolic, what else then for total time interval  t1   t2 the
it can be ?
 x 0 
average velocity is zero    0 (as
12. Name a situation where the speed of an object  t t 
is constant while the velocity is not. displacement is zero) but for 1st or 2nd time
Assess Yourself x
intervals the average velocity is not zero is .
1. A body is under constant acceleration. In its t
journey can the body move opposite to the The average velocity of the particle during a
direction of acceleration ? Give an example. time interval D t is equal to the slope of the
A. Yes ,A body projected vertically upwards straight line joining initial and final points on
the position - time graph.
2. What should be the angle of projection for a
projectile to cover the maximum range ? 6. What is the constant physical quantity that
influences the motion of a projectile ?
A. 45°
A. acceleration due to gravity
3. Under what conditions a heavy metal ball and
a feather, fall simultaneously, when they are 7. Describe the motion of a body having horizontal
dropped freely. motion with constant velocity and vertical
motion with constant acceleration ?
A. In the absence of resistive forces in vacuum
A. Projectile motion
Ex: In vacuum
8. From the top of a tower stone is dropped, while
4. Under what conditions is the magnitude of the another is thrown horizontally from the same
average velocity of a particle moving in one point at the same time. Which stone will strike
dimension smaller than the average speed over the ground first ?
same time interval ?
A. Both will strike the ground at the same time.
A. If the particle moves along a line without
9. The acceleration due to gravity is always
changing the direction, the magnitude of downward i.e., along the negative y direction.
average velocity and average speed are the Can we choose this direction as the positive
same. When change in the direction occurs direction for the acceleration due to gravity ?
displacement would be smaller than the distance,
hence average velocity would be smaller than A. Yes, the direction can be taken positive for the
'g' when the case is free fall of a body.
the average speed.
10. Can an object accelerate if its speed is constant?
5. Is it possible that the average velocity for some
interval may be zero although the average A. Yes, an object moving along a curved path with
velocity for a shorter interval included in the constant speed has varying velocity because its
first interval is not zero ? direction of velocity changes from point to point
along the trajectory.
A. Yes. If a particle moves along a straight line with
constant acceleration 1st in one direction say in 11. If the distance travelled by a particle moving
+ x direction for some interval of time  t1 , then with uniform acceleration along a straight line
it reverses its direction and moves for another is proportional to the square of the time taken,
time interval  t2 and reaches the same point what is its initial velocity ?
A. Zero
12. A person leaves his house by a cycle and returns 1) The car is stationary
to his house after travelling 25km in 2 hours. 2) The car is moving with a constant velocity V
What is his displacement ?
3) The car moves with acceleration A
A. Zero
A. In all these cases the body will have same time
13. Can a body have uniform speed and still variable of descent. The motion of the car only affects
velocity ? the magnitude of the horizontal components of
A. Yes, in case of uniform circular motion, the the velocity and the acceleration of the body
magnitude of velocity is constant, but its but does not affect the nature of its motion along
direction changes from point to point. the vertical direction.
14. Can an object accelerate if its velocity is constant 18. What is the nature of a velocity – time graph for
? a body projected vertically upwards ?
A. No, if the velocity is constant, there is no change
in the velocity hence acceleration is zero.
velocity
x
15. Can a particle have a constant velocity and A.
varying speed ? time
A. No, if velocity is constant, the magnitude of the

instantaneous velocity, i.e., speed, is constant. 19. A body moving on a circular parth of radius r is
having a velocity u at a particular point. What
16. Can a body possess velocity at the same time is the change in the velocity when it completes
both in vertical and horizontal directions ? Give a semi circle ? A.
illustration.
u   u   2u
A. Yes, in case of oblique projection or horizontal
projection at angles less than 90°, a body possess 20. When is the average velocity of a body equal to
its instantaneous velocity ?
bo th horizontal and vertical velo cities
simultaneously. A. When the body moves with uniform velocity
Eg : A bag thrown from a moving aeroplane. 21. How much force is acting on a body floating in
air ?
17. A body is dropped freely from the window of a
railway car. will the time of free fall be equal if A. Zero

Kinematics (Integreated) (Aumr Garu) (124-177)

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KINEMATICS PHYSICS - I A

h Straight line motion h Oblique projectile

4.1 INTRODUCTION (numerical value) and no direction. whereas, the

equation position is a linear function of time. Hence

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AKASH MULTIMEDIA 126

4.15 INSTANTANEOUS VELOCITY Note 4.8 : In uniform motion, the instantaneous

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4.22(i) DISPLACEMENT – TIME GRAPHS (S - T GRAPHS)

S i) A straight line parallel to x-axis.

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4.22(ii) VELOCITY – TIME GRAPHS (V – T GRAPHS) :

Context Shape of graph Comment on shape of graph

v As v = 0, the graph is a straight line

y i) A straight line with negative slope.

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4.22(iii) ACCELERATION - TIME GRAPHS (A – T GRAPHS) :

a0 The graph is a straight line parallel to

2) Particle with non-uniform a

Displacement = (Average Velocity) (time) SOLVED PROBLEMS BASED ON

AKASH MULTIMEDIA 132

Problem : 4.5 Problem : 4.8

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20´ 20 400 Problem : 4.13

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Average velocity for the first half distance = = v0 Problem : 4.18

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*Problem : 4.25 (b) Displacement = area under the v-t graph

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A displacement = (average velocity) x (time )

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*Problem : 4.30 Vav 

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Total change in velocity ds

The velocity time graph of a moving object is given

80 The corresponding v–t graph will be

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AKASH MULTIMEDIA 141

The motion of a particle along a straight line is de- V= V02 - ky 2

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Problem : 4.57 4.25 EQUATIONS OF MOTION FOR FREELY

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(ii) Time of ascent (ta) : iv) Time of flight ( tf ) :

For a body projected vertically upwards, s v

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Note 4.21: In the case of a vertically projected body, Note 4.29:

Þ both the balls fall simulateneously in the u 2 u 2 3u 2

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1 i,e . at P, g = 0 Þ v = cons tan t

Application : 4.15 Application : 4.16

Step III:- Further distance it can go up after t t1 sec

Similarly the same formula is applicable even 1 2

1 c) A body dropped from the top tower takes

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Problem : 4.64 Problem : 4.66

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