PHYSICS
CHAPTER 2

sx

sy

g

a
x
CHAPTER 2:
Kinematics of Linear Motion
(5 hours)

vy

vx

ay
1
PHYSICS
CHAPTER 2
2.0 Kinematics of Linear motion


is defined as the studies of motion of an objects without
considering the effects that produce the motion.
There are two types of motion:
 Linear or straight line motion (1-D)
 with constant (uniform) velocity
 with constant (uniform) acceleration, e.g. free fall motion
 Projectile motion (2-D)
 x-component (horizontal)
 y-component (vertical)
2
PHYSICS
CHAPTER 2
Learning Outcomes :
2.1 Linear Motion (1 hour)
At the end of this chapter, students should be able to:

Define and distinguish between
 Distance and displacement
 Speed and velocity
 Instantaneous velocity, average velocity and uniform
velocity
 Instantaneous acceleration, average acceleration and
uniform acceleration,

Sketch graphs of displacement-time, velocity-time and
acceleration-time.

Determine the distance travelled, displacement, velocity
and acceleration from appropriate graphs.
3
PHYSICS
CHAPTER 2
2.1. Linear motion (1-D)
2.1.1. Distance, d



scalar quantity.
is defined as the length of actual path between two points.
For example :
Q
P

The length of the path from P to Q is 25 cm.
4
PHYSICS
2.1.2



CHAPTER 2

Displacement, s
vector quantity.
is defined as the distance between initial point and final
point in a straight line.
The S.I. unit of displacement is metre (m).
Example 2.1 :
An object P moves 30 m to the east after that 15 m to the south
and finally moves 40 m to west. Determine the displacement of P
N
relative to the original position.
Solution :
O
W

P
30 m
E
15 m

10 m
S
30 m
5
PHYSICS
CHAPTER 2
The magnitude of the displacement is given by
OP  15 2  10 2  18 m
and its direction is
1  15 
θ  tan    56  from west to south
 10 
2.1.3 Speed, v



is defined as the rate of change of distance.
scalar quantity.
Equation:
change of distance
speed 
timeinterval
Δd
v
Δt
6
PHYSICS
2.1.4



Velocity, v
CHAPTER 2
is a vector quantity.
The S.I. unit for velocity is m s-1.
Average velocity, vav
 is defined as the rate of change of displacement.
 Equation:
change of displaceme nt
vav 
time interval
vav
s2  s1

t 2  t1
Δs
vav 
Δt

Its direction is in the same direction of the change in
displacement.
7
PHYSICS
CHAPTER 2
Instantaneous velocity, v
 is defined as the instantaneous rate of change of
displacement.
 Equation:
limit s
v
t  0 t
ds
v
dt

An object moves in a uniform velocity when
ds
 constant
dt
and the instantaneous velocity equals to the average velocity
at any time.
8
PHYSICS
CHAPTER 2
s
s1
Q
The gradient of the tangent to the curve at point Q
= the instantaneous velocity at time, t = t1
0

t
t1
Therefore
Gradient of s-t graph = velocity
9
PHYSICS
2.1.5



Acceleration, a
CHAPTER 2
vector quantity.
The S.I. unit for acceleration is m s-2.
Average acceleration, aav
 is defined as the rate of change of velocity.
change of velocity
 Equation:
aav 
time interval
v2  v1
aav 
t 2  t1
Δv
aav 
Δt


Its direction is in the same direction of motion.
The acceleration of an object is uniform when the magnitude of
velocity changes at a constant rate and along fixed direction. 10
PHYSICS
CHAPTER 2
Instantaneous acceleration, a
 is defined as the instantaneous rate of change of velocity.
 Equation:
limit v
a
t  0 t
2
dv d s
a
 2
dt dt

An object moves in a uniform acceleration when
dv
 constant
dt
and the instantaneous acceleration equals to the average
acceleration at any time.
11
PHYSICS
v
CHAPTER 2
Deceleration, a
 is a negative acceleration.
 The object is slowing down meaning the speed of the object
decreases with time.
Q
v1
The gradient of the tangent to the curve at point Q
= the instantaneous acceleration at time, t = t1
0
t1

t
Therefore
Gradient of v-t graph = acceleration
12
PHYSICS
2.1.6
CHAPTER 2
Graphical methods
Displacement against time graph (s-t)
s
s
Gradient increases
with time
Gradient = constant
t
0
s (a) Uniform velocity
t
0
(b) The velocity increases with time
Q
(c)
P
R
Gradient at point R is negative.
Gradient at point Q is zero.
0
t
The direction of
velocity is changing.
The velocity is zero.
13
PHYSICS
CHAPTER 2
Velocity versus time graph (v-t)
v
v
Uniform velocity
v
Uniform
acceleration
B
C
A
0
t1 (a) t2
t
0
t1
(b) t2
t
0
t1
t
t2 (c)
Area under the v-t graph = displacement



The gradient at point A is positive – a > 0(speeding up)
The gradient at point B is zero – a= 0
The gradient at point C is negative – a < 0(slowing down)
14
PHYSICS

CHAPTER 2
From the equation of instantaneous velocity,
ds
v
dt
ds

vdt
 
Therefore
t2
s   vdt
t1
s  shaded area under the v  t graph
Simulation 2.1
Simulation 2.2
Simulation 2.3
15
PHYSICS
CHAPTER 2
Example 2.2 :
A toy train moves slowly along a straight track according to the
displacement, s against time, t graph in Figure 2.1.
s (cm)
10
8
6
4
2
Figure 2.1 0
2 4 6 8 10 12 14
t (s)
a. Explain qualitatively the motion of the toy train.
b. Sketch a velocity (cm s-1) against time (s) graph.
c. Determine the average velocity for the whole journey.
d. Calculate the instantaneous velocity at t = 12 s.
e. Determine the distance travelled by the toy train.
16
PHYSICS
CHAPTER 2
Solution :
a. 0 to 6 s
: The train moves at a constant velocity of
6 to 10 s : The train stops.
10 to 14 s : The train moves in the same direction at a
constant velocity of
b.
v (cm s1)
1.50
0.68
0
2
4
6
8
10 12 14
t (s)
17
PHYSICS
CHAPTER 2
Solution :
vav
c.
d.
s2  s1

t2  t1
v  averagevelocityfrom10 s to14 s
s2  s1
v
t 2  t1
e. The distance travelled by the toy train is 10 cm.
18
PHYSICS
CHAPTER 2
Example 2.3 :
A velocity-time (v-t) graph in Figure 2.2 shows the motion of a lift.
v (m s 1)
4
2
0
-2
5
10 15
20 25 30 35 40 45
50 t (s)
-4
Figure 2.2
a. Describe qualitatively the motion of the lift.
b. Sketch a graph of acceleration (m s2) against time (s).
c. Determine the total distance travelled by the lift and its
displacement.
d. Calculate the average acceleration between 20 s to 40 s.
19
PHYSICS
CHAPTER 2
Solution :
a. 0 to 5 s
: Lift moves upward from rest with a constant
acceleration of
5 to 15 s : The velocity of the lift increases from 2 m s1 to
4 m s1 but the acceleration decreasing to
15 to 20 s
20 to 25 s
25 to 30 s
30 to 35 s
:
:
:
:
Lift moving with constant velocity of
Lift decelerates at a constant rate of
Lift at rest or stationary.
Lift moves downward with a constant acceleration
of
35 to 40 s : Lift moving downward with constant velocity
of
40 to 50 s : Lift decelerates at a constant rate of
and comes to rest.
20
PHYSICS
CHAPTER 2
Solution :
2
b. a (m s )
0.8
0.6
0.4
0.2
0
-0.2
5
10 15
20 25 30 35 40 45
50
t (s)
-0.4
-0.6
-0.8
21
PHYSICS
CHAPTER 2
Solution :
c. i. v (m s
1)
4
2
0
-2
A1
5
A2
10 15
A3
20 25 30 A35 40 45
4
A5
50 t (s)
-4
T otaldistance area under thegraph of v-t
 A1  A2  A3  A4  A5
22
PHYSICS
CHAPTER 2
Solution :
c. ii. Displacement  area under
thegraph of v-t
 A1  A2  A3  A4  A5
d.
v2  v1
aav 
t 2  t1
23
PHYSICS
CHAPTER 2
Exercise 2.1 :
1. Figure 2.3 shows a velocity versus time graph for an object
constrained to move along a line. The positive direction is to
the right.
Figure 2.3
a. Describe the motion of the object in 10 s.
b. Sketch a graph of acceleration (m s-2) against time (s) for
the whole journey.
c. Calculate the displacement of the object in 10 s.
24
ANS. : 6 m
PHYSICS
CHAPTER 2
Exercise 2.1 :
2. A train pulls out of a station and accelerates steadily for 20 s
until its velocity reaches 8 m s1. It then travels at a constant
velocity for 100 s, then it decelerates steadily to rest in a further
time of 30 s.
a. Sketch a velocity-time graph for the journey.
b. Calculate the acceleration and the distance travelled in
each part of the journey.
c. Calculate the average velocity for the journey.
Physics For Advanced Level, 4th edition, Jim Breithaupt, Nelson
Thornes, pg.15, no. 1.11
ANS. : 0.4 m s2,0 m s2,-0.267 m s2, 80 m, 800 m, 120 m;
6.67 m s1.
25
PHYSICS
CHAPTER 2
Learning Outcome :
2.2 Uniformly accelerated motion (1 hour)
At the end of this chapter, students should be able to:

Derive and apply equations of motion with uniform
acceleration:
v  u  at
1 2
s  ut  at
2
2
2
v  u  2as
26
PHYSICS
CHAPTER 2
2.2. Uniformly accelerated motion

From the definition of average acceleration, uniform (constant)
acceleration is given by
vu
a
t
v  u  at
(1)
where v : final velocity
u : initial velocity
a : uniform (constant) acceleration
t : time
27
PHYSICS

CHAPTER 2
From equation (1), the velocity-time graph is shown in Figure
velocity
2.4 :
v
u
Figure 2.4
t
0
time

From the graph,

The displacement after time, s = shaded area under the
graph
= the area of trapezium
Hence,
1
s  u  v t
2
(2)
28
PHYSICS

CHAPTER 2
By substituting eq. (1) into eq. (2) thus
1
s  u  u  at t
2
1 2
s  ut  at
2


From eq. (1),
From eq. (2),
v  u   at
2s
v  u  
(3)
multiply
t
 2s 
v  u v  u    at
 t 
v  u  2as
2
2
(4)
29
PHYSICS

CHAPTER 2
Notes:
 equations (1) – (4) can be used if the motion in a straight
line with constant acceleration.

For a body moving at constant velocity, ( a = 0) the
equations (1) and (4) become
v u
Therefore the equations (2) and (3) can be written as
s  vt
constant velocity
30
PHYSICS
CHAPTER 2
Example 2.4 :
A plane on a runway accelerates from rest and must attain takeoff
speed of 148 m s1 before reaching the end of the runway. The
plane’s acceleration is uniform along the runway and of value
914 cm s2. Calculate
a. the minimum length of the runway required by the plane to
takeoff.
b. the time taken for the plane cover the length in (a).
Solution :
a  9.14 m s 2
v  148 m s 1
u0
a. Use
v  u  2as
2
2
s?
t ?
31
PHYSICS
CHAPTER 2
Solution :
b. By using the equation of linear motion,
v  u  at
OR
1 2
s  ut  at
2
32
PHYSICS
CHAPTER 2
Example 2.5 :
A bus travelling steadily at 30 m s1 along a straight road passes a
stationary car which, 5 s later, begins to move with a uniform
acceleration of 2 m s2 in the same direction as the bus. Determine
a. the time taken for the car to acquire the same velocity as the
bus,
b. the distance travelled by the car when it is level with the bus.
1
2
Solution : vb  30 m s  constant ; uc  0; ac  2 ms
1
a. Given vc  vb  30 m s
Use vc  uc  ac tc
33
PHYSICS
CHAPTER 2
b.
b
c
vb  30 m s 1
b
b
ac  2 m s 2
uc  0
tb  0 s
vb
vb
c
tb  5 s
tb  t
sc  sb
From the diagram,
tb  t; tc  t  5
sc  sb
1
2
uc tc  ac tc  vbtb
2
Therefore
sc  vb t
34
PHYSICS
CHAPTER 2
Example 2.6 :
A particle moves along horizontal line according to the equation
s  t 2  2t  3
Where s is displacement in meters and t is time in seconds.
At time, t = 3 s, determine
a. the displacement of the particle,
b. Its velocity, and
c. Its acceleration.
Solution :
a. t =3 s ;
s  t  2t  3
2
35
PHYSICS
CHAPTER 2
Solution :
b. Instantaneous velocity at t = 3 s,
ds
v
dt
Use


d 2
v
t  2t  3
dt
Thus
36
PHYSICS
CHAPTER 2
Solution :
c. Instantaneous acceleration at t = 3 s,
Use
dv
a
dt
Hence
37
PHYSICS
CHAPTER 2
Exercise 2.2 :
1. A speedboat moving at 30.0 m s-1 approaches stationary
buoy marker 100 m ahead. The pilot slows the boat with a
constant acceleration of -3.50 m s-2 by reducing the throttle.
a. How long does it take the boat to reach the buoy?
b. What is the velocity of the boat when it reaches the buoy?
No. 23,pg. 51,Physics for scientists and engineers with
modern physics, Serway & Jewett,6th edition.
ANS. : 4.53 s; 14.1 m s1
2. An unmarked police car travelling a constant 95 km h-1 is
passed by a speeder traveling 140 km h-1. Precisely 1.00 s
after the speeder passes, the policemen steps on the
accelerator; if the police car’s acceleration is 2.00 m s-2, how
much time passes before the police car overtakes the
speeder (assumed moving at constant speed)?
No. 44, pg. 41,Physics for scientists and engineers with
modern physics, Douglas C. Giancoli,3rd edition.
38
ANS. : 14.4 s
PHYSICS
CHAPTER 2
Exercise 2.2 :
3. A car traveling 90 km h-1 is 100 m behind a truck traveling 75
km h-1. Assuming both vehicles moving at constant velocity,
calculate the time taken for the car to reach the truck.
No. 15, pg. 39,Physics for scientists and engineers with
modern physics, Douglas C. Giancoli,3rd edition.
ANS. : 24 s
4. A car driver, travelling in his car at a constant velocity of
8
m s-1, sees a dog walking across the road 30 m ahead. The
driver’s reaction time is 0.2 s, and the brakes are capable of
producing a deceleration of 1.2 m s-2. Calculate the distance
from where the car stops to where the dog is crossing,
assuming the driver reacts and brakes as quickly as
possible.
ANS. : 1.73 m
39
PHYSICS
Learning Outcome :
CHAPTER 2
2.3 Freely falling bodies (1 hour)
At the end of this chapter, students should be able to:

Describe and use equations for freely falling bodies.

For upward and downward motion, use
a = g = 9.81 m s2
40
PHYSICS
2.3


CHAPTER 2
Freely falling bodies
is defined as the vertical motion of a body at constant
acceleration, g under gravitational field without air
resistance.
In the earth’s gravitational field, the constant acceleration
 known as acceleration due to gravity or free-fall
acceleration or gravitational acceleration.
the value is g = 9.81 m s2
 the direction is towards the centre of the earth
(downward).
Note:
 In solving any problem involves freely falling bodies or free
fall motion, the assumption made is ignore the air
resistance.


41
PHYSICS

CHAPTER 2
Sign convention:
+
From the sign convention
thus,
+
-
a  g

Table 2.1 shows the equations of linear motion and freely falling
bodies.
Table 2.1
Linear motion
Freely falling bodies
v  u  at
2
2
v  u  2as
1 2
s  ut  at
2
v 2  u 2  2 gs
1 2
s  ut  gt
2
v  u  gt
42
PHYSICS

CHAPTER 2
An example of freely falling body is the motion of a ball thrown
vertically upwards with initial velocity, u as shown in Figure 2.5.
velocity = 0
vu
H
Figure 2.5
u
v

Assuming air resistance is negligible, the acceleration of the ball, a
= g when the ball moves upward and its velocity decreases to
zero when the ball reaches the maximum height, H.
43
PHYSICS
CHAPTER 2
s
The graphs in Figure 2.6 show
the motion of the ball moves H
up and down.
Derivation of equations
 At the maximum height or
displacement, H where t = t1, 0
v
its velocity,

hence
v0
u
v  u  gt
0
0  u  gt1
therefore the time taken for the
ball reaches H,
Simulation 2.4
Figure 2.6
t1
2t1
t1
2t1
t1
2t1
t
t
u
a
0
u
t1 
g
v =0
g
t
44
PHYSICS
CHAPTER 2
To calculate the maximum height or displacement, H:
1 2
use either

s  ut1 
2
gt1
OR
Where s = H
v 2  u 2  2 gs
0  u  2 gH
2
maximum height,

u2
H
2g
Another form of freely falling bodies expressions are
v  u  gt
2
2
v  u  2 gs
1 2
s  ut  gt
2
v y  u y  gt
v y2  u y2  2 gs y
1 2
s y  u y t  gt
2
45
PHYSICS
CHAPTER 2
Example 2.7 :
A ball is thrown from the top of a building is given an initial velocity
of 10.0 m s1 straight upward. The building is 30.0 m high and the
ball just misses the edge of the roof on its way down, as shown in
B
figure 2.7. Calculate
a. the maximum height of the stone from point A.
b. the time taken from point A to C.
u =10.0 m s1
c. the time taken from point A to D.
C
A
d. the velocity of the ball when it reaches point D.
(Given g = 9.81 m s2)
30.0 m
Figure 2.7
46
D
PHYSICS
CHAPTER 2
Solution :
a. At the maximum height, H, vy = 0 and u = uy = 10.0 m s1 thus
B
v 2y  u 2y  2 gsy
2
0  10.0   29.81H
u
C
A
H  5.10 m
b. From point A to C, the vertical displacement, sy= 0 m thus
1 2
s y  u y t  gt
2
30.0 m
D
47
PHYSICS
CHAPTER 2
Solution :
c. From point A to D, the vertical displacement, sy= 30.0 m thus
B
u
C
A
1 2
s y  u y t  gt
2 1
 30.0  10.0 t  9.81t 2
2
2
4.91t  10.0t  30.0  0
a
30.0 m
b
By using
t  3.69 s
D
c
OR
 1.66 s
Time don’t
have
negative
value.
48
PHYSICS
CHAPTER 2
Solution :
d. Time taken from A to D is t = 3.69 s thus
B
v y  u y  gt
u
C
A
30.0 m
OR
From A to D, sy = 30.0 m
2
2
v y  u y  2 gsy
Therefore the ball’s velocity at D is
D
49
PHYSICS
CHAPTER 2
Example 2.8 :
A book is dropped 150 m from the ground. Determine
a. the time taken for the book reaches the ground.
b. the velocity of the book when it reaches the ground.
(Given g = 9.81 m s-2)
Solution :
uy = 0 m s1
a. The vertical displacement is
sy = 150 m
Hence
s y  150 m
150 m
1 2
s y  u y t  gt
2
50
PHYSICS
CHAPTER 2
Solution :
b. The book’s velocity is given by
uy  0
v y  u y  gt
OR
s y  150 m
v y  u y  2 gsy
2
2
vy  ?
Therefore the book’s velocity is
51
PHYSICS
CHAPTER 2
Exercise 2.3 :
1. A ball is thrown directly downward, with an initial speed of
8.00 m s1, from a height of 30.0 m. Calculate
a. the time taken for the ball to strike the ground,
b. the ball’s speed when it reaches the ground.
ANS. : 1.79 s; 25.6 m s1
2. A falling stone takes 0.30 s to travel past a window 2.2 m tall
as shown in Figure 2.8.
2.2 m
to travel this
distance took
0.30 s
Figure 2.8
From what height above the top of the windows did the stone
fall?
52
ANS. : 1.75 m
PHYSICS
CHAPTER 2
Exercise 2.3 :
1. A ball is thrown directly downward, with an initial speed of
8.00 m s1, from a height of 30.0 m. Calculate
a. the time taken for the ball to strike the ground,
b. the ball’s speed when it reaches the ground.
ANS. : 1.79 s; 25.6 m s1
2. A falling stone takes 0.30 s to travel past a window 2.2 m tall
as shown in Figure 2.8.
2.2 m
to travel this
distance took
0.30 s
Figure 2.8
From what height above the top of the windows did the stone
fall?
53
ANS. : 1.75 m
PHYSICS
Learning Outcomes :
CHAPTER 2
2.4 Projectile motion (2 hours)
At the end of this chapter, students should be able to:

Describe and use equations for projectile,
u x  u cos θ
u y  u sin θ
ax  0
ay  g

Calculate: time of flight, maximum height, range and
maximum range, instantaneous position and velocity.
54
PHYSICS
CHAPTER 2
2.4. Projectile motion

A projectile motion consists of two components:
 vertical component (y-comp.)
motion under constant acceleration, ay= g
horizontal component (x-comp.)


motion with constant velocity thus ax= 0
The path followed by a projectile is called trajectory is shown in
Figure 2.9. y


v1
v1y
P
Simulation 2.5
uy
A
Figure 2.9
B
v
1
v1x
u
sy=H
Q v
2x
v2y

ux
2
v2
C
t1
sx= R
t2
x
55
PHYSICS

CHAPTER 2
From Figure 2.9,
 The x-component of velocity along AC (horizontal) at any
point is constant,
u x  u cos θ

The y-component (vertical) of velocity varies from one
point to another point along AC.
but the y-component of the initial velocity is given by
u y  u sin θ
56
PHYSICS

CHAPTER 2
Table 2.2 shows the x and y-components, magnitude and
direction of velocities at points P and Q.
Velocity
Point P
x-comp.
v1x  u x  u cos θ
v2 x  u x  u cos θ
y-comp.
v1y  u y  gt1
v2 y  u y  gt2
magnitude
direction
v1 
v1x 
2
Point Q
 
 v1 y
2
v
1  1 y


θ1  tan 
 v1x 
v2 
v2 x 
2
 
 v2 y
2
v
1  2 y


θ2  tan 
 v2 x 
Table 2.2
57
PHYSICS
CHAPTER 2
2.4.1 Maximum height, H

The ball reaches the highest point at point B at velocity, v
where
 x-component of the velocity, v  v  u  u cos θ
x
x
 y-component of the velocity, v y  0
 y-component of the displacement, s y  H

Use
v  u  2 gs y
2
y
2
y
0  u sin    2 gH
2
u sin 
H
2g
2
2
58
PHYSICS
CHAPTER 2
2.4.2 Time taken to reach maximum height, t’
 At maximum height, H
 Time, t = t’ and vy= 0

Use
v y  u y  gt
0  u sin    gt '
u sin 
t ' 
g
2.4.3 Flight time, t (from point A to point C)
t  2t '
2u sin θ
t 
g
59
PHYSICS
CHAPTER 2
2.4.4 Horizontal range, R and value of R maximum

Since the x-component for velocity along AC is constant hence

From the displacement formula with uniform velocity, thus the
x-component of displacement along AC is
u x  vx  u cos 
sx  u xt and sx  R
R  u cos t 
 2u sin  

R  u cos  
g


2
u
R  2 sin  cos  
g
60
PHYSICS

CHAPTER 2
From the trigonometry identity,
thus
sin 2  2 sin  cos
2
u
R
sin 2
g

The value of R maximum when  = 45 and sin 2 = 1
therefore
2
Rmax
u

g
Simulation 2.6
61
PHYSICS
CHAPTER 2
2.4.5 Horizontal projectile

Figure 2.10 shows a ball bearing rolling off the end of a table
with an initial velocity, u in the horizontal direction.
u
u
vx
vy
h
Figure 2.10

A
v
B
x
Horizontal component along path AB.
velocity, u x  u  vx  constant
displaceme nt, s x  x

Vertical component along path AB.
initial velocity, u y  0
displaceme nt, s y  h
Simulation 2.7
62
PHYSICS
CHAPTER 2
Time taken for the ball to reach the floor (point B), t
 By using the equation of freely falling bodies,
1 2
s y  u y t  gt
2
1 2
 h  0  gt
2
2h
t
g
Horizontal displacement, x
 Use condition below :
The time taken for the
ball free fall to point A
Figure 2.11
=
The time taken for the
ball to reach point B
(Refer to Figure 2.11)
63
PHYSICS

CHAPTER 2
Since the x-component of velocity along AB is constant, thus
the horizontal displacement, x
s x  u x t and s x  x
 2h 

x  u 

g



Note :
 In solving any calculation problem about projectile motion,
the air resistance is negligible.
64
PHYSICS
CHAPTER 2
Example 2.9 :
y
u
Figure 2.12 O
H
 = 60.0
P
R
v1y
Figure 2.12 shows a ball thrown by superman
with an initial speed, u = 200 m s-1 and makes an
angle,  = 60.0 to the horizontal. Determine
a. the position of the ball, and the magnitude and
direction of its velocity, when t = 2.0 s.
v1x
v1
Q
v2y
x
v2x
v2
65
PHYSICS
CHAPTER 2
b. the time taken for the ball reaches the maximum height, H and
calculate the value of H.
c. the horizontal range, R
d. the magnitude and direction of its velocity when the ball
reaches the ground (point P).
e. the position of the ball, and the magnitude and direction of its
velocity at point Q if the ball was hit from a flat-topped hill with
the time at point Q is 45.0 s.
(Given g = 9.81 m s-2)
Solution :
The component of Initial velocity :
1
ux  200cos60.0  100m s
u y  200sin 60.0  173m s1

66
PHYSICS
CHAPTER 2
Solution :
a. i. position of the ball when t = 2.0 s ,
Horizontal component :
sx  u xt
Vertical component :
1 2
s y  u y t  gt
2
therefore the position of the ball is (200 m, 326 m)
67
PHYSICS
CHAPTER 2
Solution :
a. ii. magnitude and direction of ball’s velocity at t = 2.0 s ,
Horizontal component :
vx  u x  100 m s 1
Vertical component :
v y  u y  gt
Magnitude,
Direction,
from positive x-axis anticlockwise
68
PHYSICS
CHAPTER 2
Solution :
b. i. At the maximum height, H :
vy  0
Thus the time taken to reach maximum height is given by
v y  u y  gt
ii. Apply
1 2
s y  u y t  gt
2
69
PHYSICS
CHAPTER 2
Solution :
c. Flight time = 2(the time taken to reach the maximum height)
t  217.6 
t  35.2 s
Hence the horizontal range, R is
sx  u xt
d.
When the ball reaches point P thus
The velocity of the ball at point P,
Horizontal component: 1x
x
Vertical component:
1y
y
sy  0
v  u  100 m s
v  u  gt
1
70
PHYSICS
CHAPTER 2
Solution :
Magnitude,
Direction,
therefore the direction of ball’s velocity is
from positive x-axis anticlockwise
e. The time taken from point O to Q is 45.0 s.
i. position of the ball when t = 45.0 s,
Horizontal component :
sx  u xt
71
PHYSICS
CHAPTER 2
Solution :
Vertical component :
1 2
s y  u y t  gt
2
therefore the position of the ball is (4500 m, 2148 m)
e. ii. magnitude and direction of ball’s velocity at t = 45.0 s ,
Horizontal component :
1
2x
x
Vertical component :
v  u  100 m s
v2 y  u y  gt
72
PHYSICS
CHAPTER 2
Solution :
Magnitude,
v2  v  v
2
2x
2
2y
 v2 y 

θ  tan 
 v2 x 
Direction,
1
therefore the direction of ball’s velocity is
from positive x-axis anticlockwise
73
PHYSICS
CHAPTER 2
Example 2.10 :
A transport plane travelling at a constant velocity of 50 m s1 at an
altitude of 300 m releases a parcel when directly above a point X
on level ground. Calculate
a. the flight time of the parcel,
b. the velocity of impact of the parcel,
c. the distance from X to the point of impact.
(Given g = 9.81 m s-2)
Solution :
u  50 m s 1
300 m
X
d
74
PHYSICS
CHAPTER 2
Solution :
The parcel’s velocity = plane’s velocity
1
thus
a.
u  50 m s
1
1
and
u

0
m
s
u x  u  50 m s
y
The vertical displacement is given by
s y  300 m
Thus the flight time of the parcel is
1 2
s y  u y t  gt
2
75
PHYSICS
CHAPTER 2
Solution :
b. The components of velocity of impact of the parcel :
1
Horizontal component : v x  u x  50 m s
Vertical component : v y  u y  gt
Magnitude,
Direction,
therefore the direction of parcel’s velocity is
from positive x-axis anticlockwise
76
PHYSICS
CHAPTER 2
Solution :
c.
Let the distance from X to the point of impact is d.
Thus the distance, d is given by
sx  u xt
77
PHYSICS
CHAPTER 2
Exercise 2.4 :
Use gravitational acceleration, g = 9.81 m s2
1. A basketball player who is 2.00 m tall is standing on the floor
10.0 m from the basket, as in Figure 2.13. If he shoots the
ball at a 40.0 angle above the horizontal, at what initial
speed must he throw so that it goes through the hoop without
striking the backboard? The basket height is 3.05 m.
Figure 2.13
ANS. : 10.7 m
s1
78
PHYSICS
CHAPTER 2
Exercise 2.4 :
2. An apple is thrown at an angle of 30 above the horizontal
from the top of a building 20 m high. Its initial speed is
40 m s1. Calculate
a. the time taken for the apple to strikes the ground,
b. the distance from the foot of the building will it strikes
the ground,
c. the maximum height reached by the apple from the
ground.
ANS. : 4.90 s; 170 m; 40.4 m
3. A stone is thrown from the top of one building toward a tall
building 50 m away. The initial velocity of the ball is 20 m s1
at 40 above the horizontal. How far above or below its
original level will the stone strike the opposite wall?
ANS. : 10.3 m below the original level.
79
PHYSICS
CHAPTER 2
THE END…
Next Chapter…
CHAPTER 3 :
Momentum and Impulse
80