GENERAL PHYSICS 1

Lesson 3 : Kinematics
• Acceleration Due to Gravity (FREE-FALL)
• Projectile Motion
 Acceleration Due to Gravity (FREE-
FALL)
FREE-FALL is a uniformly accelerated motion.
 One dimensional vertical motion;
 A free-falling object moves with a uniform
acceleration due to gravity (9.8 );
 The direction of acceleration due to gravity is
toward the center of the Earth (downward);
 The distance covered by a free falling object
from rest can be calculated by the equation;
d= g
 with no air resistance
 Acceleration Due to Gravity (FREE-
FALL)
 When an object is being dropped:
• the velocity and displacement is negative
(↓);
• the initial velocity () is 0
• the final velocity () is not zero
 When an object is being thrown up:
• the velocity and displacement is positive
(↑);
• the initial velocity () is not zero
• the final velocity () is 0 (at the maximum
 Kinematic Equations for Free-fall

• Vfy = Viy + gt
• Vfy = Viy + 2gy
2 2 Derive the applicable formula
based on the given

• y = V iy t + gt 2
 Example
Problem # 1:
A coin was tossed upward with the initial velocity of 5 m/s.
a) Find the highest point reached by the coin. (y=?)
b) Assuming the coin was caught by the hand that tossed it, Find the total time of flight. (t=?)
Given:
Viy = 5 m/s
Vfy = 0 m/s
g = - 9.8 m/s2
y=?
t = ?s
Solve for y using : Vfy2 = Viy2 + 2gy
 a. Find the highest point reached by the coin. (y=?)

Derivation: Solution:

Vfy2 = Viy2 + 2gy

-1 (-2gy) = (-Vfy2 + Viy2)
-1
=
1.28 m
 b. Assuming the coin was caught by the hand
that tossed it, Find the total time of flight. (t=?)

t=
Derivation: Solution:

Vfy = Viy + gt
-1(- gt ) = (-Vfy + Viy) -1 t =
gt = Vfy - Viy t = 0.51 s → It is just the time value of the upward (↑)

=
movement which is just half of the total time
to reach the hand back. It must be multiplied

Total time of flight = 2 x 0.51 s

by 2 in order to get the total time of flight

t=
= 1.02 s
 c. Supposed the coin was caught after 0.6 s, what
is the distance from its starting point?
Given: Solution:
t = 0.60 s y = (5 m/s) (0.60 s) + (-9.8m/s2)
Viy = 5 m/s (0.60 s)2
y = 3 m + (-9.8m/s2) (0.36 s2)
Vfy = 0 m/s
y = 3 m + (-3.53 m)

y = Viy t +
y =?
y = 3 m + (-1.77 m)

gt2
y = 1.24 m
 d. How fast was the coin moving when it was
caught? (Vfy= ?)
Given: Solution:
t = 0.60 s
Vfy = (5 m/s) + (-9.8m/s2) (0.60 s)
Viy = 5 m/s
g = -9.8 m/s2 Vfy = (5 m/s) + (-5.88 m/s)
Vfy = ?
Vfy = Viy + gt Vfy = - 0.88 m/s
 PROJECTILE MOTION
A projectile is a body launched at
an angle following a curvature
path (parabolic) called trajectory.
It has two components:
• Vertical (y) component is affected
by the downward pull of gravity.
• Horizontal (x) component is
influenced by inertia
 PROJECTILE MOTION
The vertical and the horizontal components are independent of each
other. Thus, they don’t have an effect on each other

2 CASES OF PROJECTILE.
• The trajectory is a half parabola

• The trajectory is a full parabola

 Half-parabola Trajectory
Vertical Motion: Horizontal Motion:
Viy = 0 m/s Vix = constant horizontal velocity
Vfy = gt dx = Vx t (Range)
dy = gt2

Range = → horizontal distance covered by a projectile

T= → usually used to get the time of full trajectory

 Example Problem for Half Trajectory
A ball is kicked near the edge of the cliff with a horizontal velocity (Vx) of 100 m/s.
a. What is the velocity upon impact if it hits the ground after 2 seconds?
b. What is the horizontal displacement of the ball?

Given:
Vx = 100 m/s
t=2s
V =?
Vfy = ?
x= ?
 Example Problem for Half Trajectory
b. Vx = → dx = Vxt
Solution:
a. Vx = 100 m/s
Vy = gt → (-9.8 m/s2) (2s) dx = (100 m/s) (2 s)
Vy = (-19.6 m/s) dx = 200 m

V=
V=
V=
V=
V=
 The Trajectory is a Full Parabola
Problem #1.
A ball is thrown with the initial velocity of 100 m/s at an angle of 30° above the horizontal.
a. How far from the throwing point will the ball attain its original level? (x-displacement)

Solution.
1. Find Vix and Viy independently.
Vix = Vi cos θ → 100 m/s cos 30° = 86.60 m/s
Viy = Vi sin θ → 100 m/s sin 30° = 50.00 m/s
 The Trajectory is a Full Parabola
2. Find the time of flight.
T = or t =
t=
t= → (use the absolute value since we are looking for the value of time )

t = 10.20 s (time of flight across the horizon)

 The Trajectory is a Full Parabola
R=
3. Find the x- displacement
x = Vixt
R=

x = (86.60 m/s) (10.20

R=

s)
R=

x = 883.32 m
R=
R = 883.70 m

use the absolute value

 Practice

1. A ball rolled at the edge of the table at the horizontal velocity of

20 m/s.
a. What is the velocity upon impact after 3 seconds?
b. How far will it reach from the table?
 Answer
Given: B. x- displacement
Vx = 20 m/s
Vy = (-9.8 m/s2) (3 s) = -29.40 s
dx = (20 m/s) (3 s)

A. Resultant Velocity dx = 60 m

VR =
VR =
VR =

VR = 35.56 m/s
 Problem #2
A ball is thrown from the top of the building 50 m away at the initial velocity of 20 m/s at 40° above the
horizon.

a. How far above or below its original level will the ball strike the opposite wall? ( y-displacement)

1st. Solve for the Vix & Viy 3rd. Find the y- displacement
y = Viyt + gt2

Vix = 20 m/s cos 40° = 15.32 m/s y = (12.86 m/s) (3.26s) + (-9.8m/s2) (3.26 s)2
y = 41.92 m + (-104.15 m)
Viy = 20 m/s sin 40° = 12.86 m/s
y = 41.92 m + (-52.08 m)
y = -10.16 m
2nd. Find the time. ( Derive the formula Vx = )

t= → = 3.26 s
 Problem #3.
3. A body is projected upward from the ground @ 50° with the initial velocity of 40 m/s.

a. How long will it take before striking the ground? (total time of flight)
b. How far from the starting point will it strike? (Range)
c. At what angle with the horizon will it strike? (θ)
d. Find the maximum height reached by the body.

Given: A. Find the total time of flight.

V = 40 m/s
T=
θ = 50° above the horizon
T=
1st. Find the Vix and Viy
T = -6.25 (use the absolute value)
Vix = 40 m/s cos 50° = 25.71
m/s 6.25 s
Viy = 40 m/s sin 50° = 30.64
m/s
 Problem #3.
B. Find the Range. R= C. Find the angle (θ).
x = Vixt R= tan-1 =
x = (25.71 m/s) (6.25 s) R= tan-1 =
x = 160.69 m
tan-1 = 50° above the
horizon
R=
R = 160.78 m
We can also use
the formula for
Range
 Problem #3.
…or we can use half of
D. Find the maximum height reached
the time and the formula in
by an object. finding y-displacement
h=
h= Given:
h= t = 6.25 s (divide 2) = 3.13 s
h= y = Viy t + ½ gt2

h = 47. 90 m y = (30.64 m/s) (3.13 s) + ½ (-9.8 m/s2) (3.13s)2

y = 47.90 m