PHYSICS

Physical quantities can be classified under two main headings

a. Vector-quantity that has both magnitude and direction (e.g. velocity, acceleration, force, momentum)
b.Scalar-is any quantity that has magnitude only (e.g. speed, pressure, temperature, energy)

Motion
Motion is an object’s change in position with respect to time.
Distance vs Displacement
Displacement is the change in position of the object with respect to a frame of reference

Distance is the total path covered by an object

Vector Tail Head

Arrow are used to represent vectors
The direction that the arrow is pointing represents the direction of the vector
The length of the arrow is proportional to the magnitude of the vector
Direction of Vector
A vector arrow (with arrowhead) is drawn in a specified direction. The vector
arrow has a head and a tail.
Vector diagrams depict a vector by use of an arrow drawn to scale in a
specific direction.
Vectors can be directed due East, due West, due South, and due North.
In cases where the direction of vector that is not due East, due West, due South, or due North, the direction of the vector
can be represented as degrees.
The direction of a vector is often expressed as an angle of rotation of the vector about its "tail" from east, west, north, or
south.

Example
The magnitude and direction of the vector is clearly labeled. In
this case, the diagram shows the magnitude is 20 m and the
direction is (30 degrees West of North).

Resultant

Resultant vector represents the sum of two or more vectors.

This can be obtained using the graphical or component method.
The horizontal component is known as the x component and the
vertical component is known as the y component.
Eastward and Northward vectors are denoted by positive
vectors.
Westward and Southward vectors are denoted by negative
vectors.

a. When two vectors point the same direction, simply add them together and follow the common direction.
b. When two vectors point in the opposite direction, simply subtract them and follow the direction of the vectors
having greater magnitude
c. For vectors acting perpendicular with each other, use the pythagorean theorem to solve for the magnitude
and tangent function for the direction.

Pythagorean Theorem
Assign variables.
VR=Resultant
VR 2 = VX 2 + VY 2
East and West lies along the Horizontal Plane thus vectors
Tangent Function
East and West is assigned as VX
opp V
tanθ = ortanθ = VY
North and South lies along the Vertical Plane thus vectors adj X
North and South is assigned as VY

Example:
A cat moves 4 m East and then turns 3 m North.Find the resultant

Resultant=?

3 m, North
θ=?

4 m, East
3 m, North= VY
4 m, East= VX

For magnitude For direction Resultant

VY 𝟓 𝐦, 𝟑𝟔. 𝟖𝟕𝟎° 𝐍𝐨𝐫𝐭𝐡 𝐨𝐟 𝐄𝐚𝐬𝐭

VR 2 = VX 2 + VY 2 tanθ =
VX
VR 2 = (4m)2 + (3m)2 3
VR 2 = 16m2 + 9m2 θ = tan− ( )
4
VR 2 = 25m2 −(
θ = tan 0.75)
√VR 2 = √25m2
𝛉 = 𝟑𝟔. 𝟖𝟕𝟎°
𝐕𝐑 = 𝟓 𝐦
Resolution-The process of determining the components of vector is called as resolution
 VX = cosθR
VX = cos(36.870°) (5 m)
VX = 0.8 (5 m)
VR=5 m? VX =4m

VY=? VY = sinθR
θ=36.870° VY = sin(36.870°) (5 m)
VY = 0.6 (5 m)
VX=? VY =3m

d.To solve the resultant of two or more vectors, use the component method.

Example:
An airplane travels 100 km north then 220 km east. It then changes the course by 300 km on a straight direction at an
angle off 30° east of north before reaching its destination.

V2= V3=300 km, 30°

East of North
220 km,
East
V1=

100 VR=?
km,
North

X component Y component
V3= 100 km,
0 km 100 km
North
V3= 220 km,
220 km 0 km
East
VX = cosθR VY = sinθR
VX = cos(30°) (300 km) VY = sin(30°) (5 km)
V3=300 km, 30°
East of North
VX = 0.866 (300 km) VY = 0.5 (300 km)
VX = 259.8 km VY = 150 km
ΣVX = 0 km + 220 km + 259.8 km ΣVX = 100 km + 0 km + 150 km
ΣVX = 479.8 km ΣVX = 250 km
 For magnitude For direction Resultant

VY 𝟓𝟒𝟏. 𝟎𝟐𝟓 𝐦, 𝟔𝟐. 𝟒𝟕𝟔°

VR 2 = ΣVX 2 + ΣVY 2 tanθ =
VX 𝐍𝐨𝐫𝐭𝐡 𝐨𝐟 𝐄𝐚𝐬𝐭
VR 2 = (479.8 km)2 + (250 km)2
479.8
VR 2 = 230208.04 km2 + 62500 km2 θ = tan− ( )
VR 2 = 292708 km2 250
θ = tan− (1.919)
√VR 2 = √292708 km2
𝛉 = 𝟔𝟐. 𝟒𝟕𝟔°
𝐕𝐑 = 𝟓𝟒𝟏. 𝟎𝟐𝟓 𝐤𝐦

Speed Vs Velocity
Speed is the distance travelled by an object over a period of time (Scalar Quantity)
𝑑
𝑣=
𝑡
Velocity is the displacement of an object over a period of time (Vector Quantity)
𝑑
𝑣=
𝑡

Derivation of Formula:
d d d = vt
v= t=
t v
Where:v = velocity/ speed t = time d = distance/ displacement
Sample Problem Sample Problem Sample Problem
A car is heading north An airplane flies with a constant speed Juancho drives his car with an average
covering a distance of 500 of 920 km/h. How far can it travel in velocity of 24 m/s to the north expressway.
meters in 20 sec. What is the 3.5 hours? How long will it take to drive 560 km on a
speed of the car? perfectly straight highway?
Given:
Given: km Given:
d = 500 m v = 920 m 1 km
hr v = 24 ( )
t = 20 s t = 3.5 hr s 1000 m
Find: Find: Convert m/s to km/hr
v =? d =? m 1 km 60 s 60 min
Solution: Solution: 24 ( )( )( )
d s 1000 m 1 min 1 hr
v= d = vt km
t km = 86.4
500 m d = (90 ) (3.5 hr) hr
v= hr d = 560 km
20 s 𝐯 = 𝟑𝟏𝟓 𝐤𝐦 Find:
𝐦 t =?
𝐯 = 𝟐𝟓
𝐬 Solution:
d
t=
v
560 km
t= km
86.4
hr
hr
t = (560 km) ( )
86.4 km
𝐭 = 𝟔. 𝟒𝟖 𝐡𝐫𝐬

Acceleration
the rate of change in velocity at a given time.
The action from a force can cause an object to move or speed up (accelerate), to slow down (decelerate), to stop, or
to change direction. Since any change in velocity is considered acceleration, it can be said that a force on
an object results in the acceleration of an object.
An moving object that changes direction is accelerating.
Objects that move in uniform circular motion have constant speeds but still accelerate because they constantly change in
direction as they go around the circular path.
Where
Vf − Vi a = acceleration Vi = ininal velocity
a= Vf = final velocity
t t = time
Sample Problem
Determine the acceleration of a coaster which moves with a velocity of 10 m/s, after 2s its velocity increases to 26 m/s.
Given:
m m
Vf = 26 Vi = 10 t=2s
s s
Find: a =?
Solution:
Vf − Vi
a=
t
m m m
26 − 10 16
s s s
a= =
2s 2s
m 1
a = 16 ( )
s 2s
𝐦
𝐚=𝟖
𝐬𝟐

Uniformly Accelerated Motion

-if a body maintains a constant change in velocity in a given time interval along a straight line, then the body is said to
have uniform acceleration.
- is a type of motion in which the velocity of an object changes by an equal amount in every equal time period.

Guide
Quantity Meaning
When Value of t
where Value of d
Starts from rest Vi=0
How fast Value of v
How long does it take Value of t
How far Value of d
Comes to a stop Vf=0

Equations
Vf + Vi at 2 Where
d=( )t d = vi t + a = acceleration
2 2 𝐹𝑜𝑟 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑚𝑜𝑡𝑖𝑜𝑛
𝑚
𝑎 = 9.8 2
𝑠
Vf = final velocity
Vf = √Vi 2 t + 2ad Vf = Vi + at Vi = initial velocity
t = time
d = distance
Sample Problem
An airplane from rest accelerates on runway at 5.50 m/s 2 for 20.25 s until it finally takes off the ground. What is the
distance covered before take off? What will be the final velocity?
Given:
m m
vi = 0 a = 5.50 t = 20.25 s
s s2
Find:
d =? vf =?

at 2 Vf = Vi + at
d = vi t + m m
2 Vf = 0 + (5.50) (20.25 s)
m s s2
d = 0 (20.25 s) m
s Vf = (5.50 2 ) (20.25 s)
m s
(5.50 s2 ) (20.25 s)2 𝐦
+ 𝐕𝐟 = 𝟏𝟏𝟏. 𝟑𝟕𝟓
2 𝐬
m or
(5.50 s2 ) 410.0625 s2
d=
2 Vf = √Vi 2 t + 2ad
2255.34375 m
d= m 2 m
2
Vf = √(0 ) (20.25 s) + 2 (5.50 ) (1127.672 m)
𝐝 = 𝟏𝟏𝟐𝟕. 𝟔𝟕𝟐 𝐦 s s2

m2
Vf = √2 (6202.196 )
s2

m2 𝐦
Vf = √12404.392 𝐕𝐟 = 𝟏𝟏𝟏. 𝟑𝟕𝟓
s2 𝐬

Sample Problem
Zed is playing with a ball on top of a building but the ball fell and hits the ground after 2.6 s, how high is the building and
what is the final velocity of the ball before it hits the ground?
Given:
m m
vi = 0 a = −9.8 t = 2.6 s
s s2
Find:
d =? vf =?
 at 2 Vf = Vi + at
d = vi t + m m
2 Vf = 0 + (9.8 ) (2.6 s)
m
)2 s s2
m (9.8 s2 ) (2.6 s m
d=0 (2.6 s) + Vf = (9.8 2 ) (2.6 s)
s 2 s
m 2 𝐦
(9.8 s2
) 6.76 s 𝐕𝐟 = 𝟐𝟓. 𝟒𝟖
d= 𝐬
2 or
66.248 m
d= Vf = √Vi 2 t + 2ad
2
𝐝 = 𝟑𝟑. 𝟏𝟐𝟒 𝐦
m 2 m
Vf = √(0 ) (2.6 s) + 2 (9.8 ) (33.124 m)
s s2

m2
Vf = √2 (324.6125 )
s2

m2 𝐦
Vf = √649.2304 𝐕𝐟 = 𝟐𝟓. 𝟒𝟖
s2 𝐬

PROJECTILE MOTION
Projectile
-is an object that is initially thrown into the air and continues to move on its trajectory acted upon by gravity.
Projectile Motion
is a form of motion in which an object called a projectile is thrown near the Earth's surface, and it moves along a curved
path under the action of gravity only.
Two balls released simultaneously from the same height. One ball is thrown sideways while the other ball is dropped from
rest.
Both balls reach the floor at the same time even though the yellow ball has travelled more distance. This is due to the fact
that acceleration due to gravity (g) is the same for both objects.
Types of Projectile Motion
-object dropped from rest is a projectile
-object that is thrown vertically upward
-object which is thrown upward at an angle to the horizontal
Components of Projectile Motion
The horizontal component can be compared to the motion of a ball as it rolls on a flat surface.
The vertical component can be compared to the motion of an object experiencing a free fall which is accelerating due to
the influence of gravity. 9.8 m/s2
Because gravity is the only force, the acceleration of a projectile is the acceleration of gravity 9.8 m/s2, down. As such,
projectiles travel along their trajectory with a constant horizontal velocity and a changing vertical velocity. A projectile has
a motion which is both horizontal and vertical at the same time.
Elements of Projectile Motion
1. Time- the entire duration while the projectile is in the trajectory(the path followed by a projectile)
2. Range- the maximum horizontal distance traveled by the projectile
3. Maximum height- is the maximum vertical displacement traveled by projectile in its trajectory
Projectile Horizontally Launched from an Elevated
Position
A projectile launched horizontally has noinitial vertical
velocity.
Thus, itsvertical motion is identical to that of adropped
object.
The downwardvelocity increases uniformly due togravity as
shown by the vectorarrows of increasing lengths.
Thehorizontal
velocity is uniform asshown by the identical horizontalvector
arrows.

Horizontal Motion Vertical Motion

 x = vix t at 2
vfx = vix (constant) y=
2
a=0 vfy = at
Where vfy 2 = viy 2 + 2ay
x = horizontal viy = 0
displacement, Where
(range) y = vertical displacement
vix = initial (height)
horizontal velocity viy = initial vertical velocity
vfx = final horizontal velocity
vfy = final vertical velocity
t = time m
a = acceleration due to gravity 9.8
s2
t = time

Resultant Velocity Time Horizontal Velocity

x
𝐯 = √𝐯𝐟𝐱 𝟐 + 𝐯𝐟𝐲 𝟐 2y vx =
t=√ t
a Where
Where t = time
t = time vx = horizontal velocity
y = vertical displacement(height) (range)x = horizontal displacement,
m
a = accelerationto gravity 9.8 2
s

Commonly Used Formula

Time Vertical Displacement/(Height) Horizontal Displacement/(Range)

at 2 x = vix t
2y y= Where
t=√ 2
a Where x = horizontal displacement,
Where y = vertical displacement (height) (range)
t = time m vix = initial horizontal velocity
a = acceleration due to gravity 9.8
y = vertical displacement s2 t = time
(height) t = time
m
a = accelerationto gravity 9.8
s2

Sample Problem
A ball rolls off of a table with a speed of 3.2 m/s. The table is 1.5 m high.a) When does the ball hit the ground?
b) How far away from the base of the table does the ball travel? c) With what speed does the ball hit the floor?
Given:
m
vix = 3.2 y = 1.5 m
s
 a) When does the ball hit the ground? c) With what speed does the ball hit the floor?
m
2y vfx = vix = 3.2
t=√ s
a m m
vfy = at = 9.8 (0.55 s) = 5.39
s2 s
2(1.5 m)
t=√ m
9.8 s2
v = √vfx 2 + vfy 2
= 𝟎. 𝟓𝟓 𝐬

b) How far away from the base of the m 2 m m2 m2 m2

= √(3.2 ) + (5.39 )2 = √10.24 2 + 29.05 2 = √39.29 2
table does the ball travel? s s s s s
𝐦
x = vix t = 𝟔. 𝟐𝟕 𝐬
m
= 3.2 (0.55 s)
s
= 𝟏. 𝟕𝟔 𝐦

If a bullet is fired with a speed of 600 m/s horizontally from a height of 48 m, how long will it take to hit the ground?
What is the range of the projectile?
Given:
m
vix = 600 y = 48 m
s
How long will it take to hit the ground? What is the range of the projectile?
x = vix t
2h m
t=√ = 600 (3.1 s)
a s
= 𝟏𝟖𝟔𝟎 𝐦
2(48 m)
t=√ m
9.8 s2

= 𝟑. 𝟏 𝐬

Projectile Launched At an Angle

Velocity has two components:
1. a constant horizontal velocity that moves in the
same direction as the launch, the acceleration of
which is zero
2. an upward positive vertical velocity component
that is decreasing in magnitude until it becomes
zero at the top of the trajectory (therefore it no
longer goes up any further). But because gravity
makes it accelerates downward at a rate 9.8 m/s2,
(therefore it stays at rest only for an instant) it will
start to descend with an increasing negative vertical
velocity until it is stopped by something.

During the flight, the horizontal velocity component Vx remains constant, while the vertical velocity component Vy varies.
On its way up, the vertical velocity decreases at a rate of 9.8 m/s2.
At its maximum height, the vertical velocity is zero.
As the projectile starts to fall down, its vertical velocity increases at a rate of 9.8 m/s2 as it reaches the ground.

Horizontal Motion Vertical Motion

 vix = vi cosθ viy = vi sinθ
x = vix t at 2
vfx = vix (constant) y = viy t +
2
a=0 vfy = viy + at
Where vfy 2 = viy 2 + 2ay
x = horizontal displacement, (range) Where
vix = initial horizontal velocity y = vertical displacement (height)
vfx = final horizontal velocity viy = initial vertical velocity
t = time vfy = final vertical velocity
θ = angle m
vi = initial velocity a = acceleration due to gravity9.8 2
s
t = time
θ = angle
Commonly Used Formula
Time Vertical Displacement Horizontal Displacement
(Height) (Range)
vfy − viy at 2 x = vix t
t= y = viy t +
a 2 Where
Where Where x = horizontal
t = time y = vertical displacement displacement,
viy =initial vertical velocity (height) (range)
vfy =final vertical velocity viy = initial vertical velocity vix = initial horizontal
a = acceleration a = acceleration due to velocity
m m t = time
to gravity − 9.8 2 gravity 9.8 2
s s
t = time (1/2 of the total time)

A bouncing ball leaves the ground with a velocity of 4.36 m/s at an angle of 81 degrees above the
horizontal.
a) How long did it take the ball to land?
b) How high did the ball bounce?
c) What was the ball's range?
Given:
m
vi = 4.36 θ = 81°
s

First, determine the horizontal and vertical components of the initial velocity.
vix = vi cosθ viy = vi sinθ
m m
= 4.36 cos81° = 4.36 sin81°
s s
m m
= 4.36 (0.16) = 4.36 (0.99)
s s
m m
= 0.70 = 4.32
s s
Analyze the horizontal motion and the vertical motion.
Horizontal Vertical
x =? y =?
m m
vix = 0.70 viy = 4.32
s s
m m
vfx = 0.70 vfy = −4.32
s s
m
a = −9.8 2
s

a) How long did it take the ball b) How high did the ball bounce?
to land? The problem statement asks for the height of the projectile at its peak.
vfy − viy This is the same as asking, "what is the vertical displacement (y) of the
t= projectile when it is halfway through its trajectory?" In other words, find y
a
m m when t = 0.44 seconds ;one-half of the total time (0.88 s).
−4.32 s − 4.32
= m
s at 2
−9.8 s2 y = viy t +
2
m m
−8.64 s m −9.8 (0.44s) 2
= s2
−9.8
m
= 4.32 (0.44 s) +
s2 s m
2
=𝟎. 𝟖𝟖 𝐬
−9.8 (0.19s 2 )
s2
= 1.90 m +
2
= 1.90 m − 0.93 m
= 𝟎. 𝟗𝟕 𝐦
c) What was the ball's range?
x = vix t
m
= 0.70 (0.88 s)
s
= 𝟎. 𝟔𝟐 𝐦