1. PHYSICS CHAPTER 2
1
CHAPTER 2:
Kinematics of Linear Motion
(5 hours)
xs

ys

xv

yv

xa

ya

g


2. 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

3. PHYSICS CHAPTER 2
Learning Outcomes :
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
2.1 Linear Motion (1 hour)

4. 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 :
 The length of the path from P to Q is 25 cm.
4
P
Q

5. PHYSICS CHAPTER 2
s

5
 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
relative to the original position.
Solution :
2.1.2 Displacement,
N
EW
S
O
P


30 m
15 m
10 m 30 m

6. PHYSICS CHAPTER 2
The magnitude of the displacement is given by
and its direction is
2.1.3 Speed, v
 is defined as the rate of change of distance.
 scalar quantity.
 Equation:
intervaltime
distanceofchange
speed
6
Δt
Δd
v 
m181015 22
OP
southwest tofrom56
10
15
tan 1 






 
θ

7. PHYSICS CHAPTER 2
v

7
intervaltime
ntdisplacemeofchange
avv
12
12
av
tt
ss
v



 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:
 Its direction is in the same direction of the change in
displacement.
2.1.4 Velocity,
Δt
Δs
vav 

8. PHYSICS CHAPTER 2
constant
dt
ds
8
t
s
0t
v




limit
Instantaneous velocity, v
 is defined as the instantaneous rate of change of
displacement.
 Equation:
 An object moves in a uniform velocity when
and the instantaneous velocity equals to the average velocity
at any time.
dt
ds
v 

9. PHYSICS CHAPTER 2
9
 Therefore
Q
s
t0
s1
t1
The gradient of the tangent to the curve at point Q
= the instantaneous velocity at time, t = t1
Gradient of s-t graph = velocity

10. PHYSICS CHAPTER 2
a

10
intervaltime
velocityofchange
ava
 vector quantity.
 The S.I. unit for acceleration is m s-2.
Average acceleration, aav
 is defined as the rate of change of velocity.
 Equation:
 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.
2.1.5 Acceleration,
12
12
av
tt
vv
a



Δt
Δv
aav 

11. PHYSICS CHAPTER 2
constant
dt
dv
11
t
v
0t
a




limit
Instantaneous acceleration, a
 is defined as the instantaneous rate of change of velocity.
 Equation:
 An object moves in a uniform acceleration when
and the instantaneous acceleration equals to the average
acceleration at any time.
2
2
dt
sd
dt
dv
a 

12. PHYSICS CHAPTER 2
12
Deceleration, a
 is a negative acceleration.
 The object is slowing down meaning the speed of the object
decreases with time.
 Therefore
v
t
Q
0
v1
t1
The gradient of the tangent to the curve at point Q
= the instantaneous acceleration at time, t = t1
Gradient of v-t graph = acceleration

13. PHYSICS CHAPTER 2
13
Displacement against time graph (s-t)
2.1.6 Graphical methods
s
t0
s
t0
(a) Uniform velocity (b) The velocity increases with time
Gradient = constant
Gradient increases
with time
(c)
s
t0
Q
R
P
The direction of
velocity is changing.
Gradient at point R is negative.
Gradient at point Q is zero.
The velocity is zero.

14. PHYSICS CHAPTER 2
14
Velocity versus time graph (v-t)
 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)
t1 t2
v
t0
(a) t2t1
v
t0
(b)
t1 t2
v
t0
(c)
Uniform velocity
Uniform
acceleration
Area under the v-t graph = displacement
B
C
A

15. PHYSICS CHAPTER 2
dt
ds
v 
15
 From the equation of instantaneous velocity,
Therefore
  vdtds

2
1
t
t
vdts
graphunder theareadedsha tvs 
Simulation 2.1 Simulation 2.2 Simulation 2.3

16. PHYSICS CHAPTER 2
16
A toy train moves slowly along a straight track according to the
displacement, s against time, t graph in Figure 2.1.
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.
Example 2.2 :
0 2 4 6 8 10 12 14 t (s)
2
4
6
8
10
s (cm)
Figure 2.1

17. PHYSICS CHAPTER 2
17
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.
0 2 4 6 8 10 12 14 t (s)
0.68
1.50
v (cm s1)

18. PHYSICS CHAPTER 2
18
Solution :
c.
d.
e. The distance travelled by the toy train is 10 cm.
12
12
tt
ss
vav



s14tos10fromvelocityaveragev
12
12
tt
ss
v




19. PHYSICS CHAPTER 2
19
A velocity-time (v-t) graph in Figure 2.2 shows the motion of a lift.
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.
Example 2.3 :
0
5 10 15 20 25 30 35 t (s)
-4
-2
2
4
v (m s 1)
Figure 2.2
40 45 50

20. PHYSICS CHAPTER 2
20
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 : Lift moving with constant velocity of
20 to 25 s : Lift decelerates at a constant rate of
25 to 30 s : Lift at rest or stationary.
30 to 35 s : 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.

21. PHYSICS CHAPTER 2
21
Solution :
b.
t (s)5 10 15 20 25 30 35 40 45 50
0
-0.4
-0.2
0.2
0.6
a (m s2)
-0.6
-0.8
0.8
0.4

22. PHYSICS CHAPTER 2
22
Solution :
c. i.
0
5 10 15 20 25 30 35 t (s)
-4
-2
2
4
v (m s 1)
40 45 50
A1
A2 A3
A4
A5
v-tofgraphunder theareadistanceTotal 
54321 AAAAA 

23. PHYSICS CHAPTER 2
23
Solution :
c. ii.
d.
v-tofgraphunder theareantDisplaceme 
54321 AAAAA 
12
12
tt
vv
aav




24. PHYSICS CHAPTER 2
24
Figure 2.3
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.
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.
ANS. : 6 m
Exercise 2.1 :

25. PHYSICS CHAPTER 2
25
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.
Exercise 2.1 :

26. PHYSICS CHAPTER 2
Learning Outcome :
At the end of this chapter, students should be able to:
 Derive and apply equations of motion with uniform
acceleration:
26
2.2 Uniformly accelerated motion (1 hour)
atuv 
2
2
1
atuts 
asuv 222


27. PHYSICS CHAPTER 2
2.2. Uniformly accelerated motion
 From the definition of average acceleration, uniform (constant)
acceleration is given by
where v : final velocity
u : initial velocity
a : uniform (constant) acceleration
t : time
27
atuv  (1)
t
uv
a



28. PHYSICS CHAPTER 2
28
 From equation (1), the velocity-time graph is shown in Figure
2.4 :
 From the graph,
The displacement after time, s = shaded area under the
graph
= the area of trapezium
 Hence,
velocity
0
v
u
timet
Figure 2.4
 tvu
2
1
s  (2)

29. PHYSICS CHAPTER 2
29
 By substituting eq. (1) into eq. (2) thus
 From eq. (1),
 From eq. (2),
  tatuus 
2
1
(3)
2
2
1
atuts 
  atuv 
 
t
s
uv
2

multiply
    at
t
s
uvuv 






2
asuv 222
 (4)

30. PHYSICS CHAPTER 2
30
 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
Therefore the equations (2) and (3) can be written as
uv 
vts  constant velocity

31. PHYSICS CHAPTER 2
31
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. Use
Example 2.4 :
?t
asuv 222

0u
?s
2
sm14.9 
a 1
sm148 
v

32. PHYSICS CHAPTER 2
32
Solution :
b. By using the equation of linear motion,
atuv 
OR
2
2
1
atuts 

33. PHYSICS CHAPTER 2
33
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.
Solution :
a. Given
Use
Example 2.5 :
21
ms20;;constantsm30 
 ccb auv
cccc tauv 
1
sm30 
 bc vv

34. PHYSICS CHAPTER 2
34
b.
From the diagram,
c
b
1
sm30 
bv
0cu
s0bt s5bt
2
sm2 
ca
b
bv
b
c
bv
ttb 
bc ss 
bc ss 
bbcccc tvtatu 
2
2
1
Therefore
tvs bc 
;ttb  5 ttc

35. PHYSICS CHAPTER 2
35
A particle moves along horizontal line according to the equation
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 ;
Example 2.6 :
322
 tts
322
 tts

36. PHYSICS CHAPTER 2
36
Solution :
b. Instantaneous velocity at t = 3 s,
Use
Thus
dt
ds
v 
 322
 tt
dt
d
v

37. PHYSICS CHAPTER 2
37
Solution :
c. Instantaneous acceleration at t = 3 s,
Use
Hence
dt
dv
a 

38. PHYSICS CHAPTER 2
38
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.
ANS. : 14.4 s
Exercise 2.2 :

39. PHYSICS CHAPTER 2
39
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
Exercise 2.2 :

40. PHYSICS CHAPTER 2
Learning Outcome :
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
2.3 Freely falling bodies (1 hour)

41. PHYSICS CHAPTER 2
2.3 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

42. PHYSICS CHAPTER 2
42
 Sign convention:
 Table 2.1 shows the equations of linear motion and freely falling
bodies.
Table 2.1
Linear motion Freely falling bodies
atuv  gtuv 
as2uv 22
 gs2uv 22

2
at
2
1
uts 
2
gt
2
1
uts 
+
- +
-
From the sign convention
thus,
ga 

43. PHYSICS CHAPTER 2
43
 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.
 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.
H
u
v
velocity = 0
Figure 2.5
uv 

44. PHYSICS CHAPTER 2
gtuv 
44
 The graphs in Figure 2.6 show
the motion of the ball moves
up and down.
Derivation of equations
 At the maximum height or
displacement, H where t = t1,
its velocity,
hence
therefore the time taken for the
ball reaches H,
Figure 2.6
t0
v
u
u
t1 2t1
t0
a
g
t1 2t1
t
s
0
H
t1 2t1
v =0
1gtu 0
0v
g
u
t1 
Simulation 2.4

45. PHYSICS CHAPTER 2
2
11 gtuts
2
1

45
 To calculate the maximum height or displacement, H:
use either
maximum height,
 Another form of freely falling bodies expressions are
gsuv 22
2
Where s = H
gHu 20 2

OR
g
u
H
2
2

gtuv 
gsuv 222

2
2
1
gtuts 
gtuv yy 
yyy gsuv 222

2
2
1
gttus yy 

46. PHYSICS CHAPTER 2
46
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
figure 2.7. Calculate
a. the maximum height of the stone from point A.
b. the time taken from point A to C.
c. the time taken from point A to D.
d. the velocity of the ball when it reaches point D.
(Given g = 9.81 m s2)
Example 2.7 :
A
B
C
D
u =10.0 m s1
30.0 m
Figure 2.7

47. PHYSICS CHAPTER 2
y
2
y
2
y gsuv 2
   H9.81210.00
2

47
Solution :
a. At the maximum height, H, vy = 0 and u = uy = 10.0 m s1 thus
b. From point A to C, the vertical displacement, sy= 0 m thus
m5.10H
2
yy gttus
2
1

A
B
C
D
u
30.0 m

48. PHYSICS CHAPTER 2
48
Solution :
c. From point A to D, the vertical displacement, sy= 30.0 m thus
By using
2
yy gttus
2
1

s3.69t
    2
tt 9.81
2
1
10.030.0 
A
B
C
D
u
30.0 m
030.010.04.91  tt2
OR s1.66
Time don’t
have
negative
value.
a b c

49. PHYSICS CHAPTER 2
gtuv yy 
49
Solution :
d. Time taken from A to D is t = 3.69 s thus
From A to D, sy = 30.0 m
Therefore the ball’s velocity at D is
A
B
C
D
u
30.0 m
OR
y
2
y
2
y gsuv 2

50. PHYSICS CHAPTER 2
50
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 :
a. The vertical displacement is
sy = 150 m
Hence
Example 2.8 :
uy = 0 m s1
150 mm150ys
2
yy gttus
2
1


51. PHYSICS CHAPTER 2
gtuv yy 
51
Solution :
b. The book’s velocity is given by
Therefore the book’s velocity is
OR
y
2
y
2
y gsuv 2
m150ys
0yu
?yv

52. PHYSICS CHAPTER 2
52
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.
From what height above the top of the windows did the stone
fall?
ANS. : 1.75 m
Exercise 2.3 :
m2.2
Figure 2.8
to travel this
distance took
0.30 s

53. PHYSICS CHAPTER 2
53
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.
From what height above the top of the windows did the stone
fall?
ANS. : 1.75 m
Exercise 2.3 :
m2.2
Figure 2.8
to travel this
distance took
0.30 s

54. PHYSICS CHAPTER 2
Learning Outcomes :
At the end of this chapter, students should be able to:
 Describe and use equations for projectile,
 Calculate: time of flight, maximum height, range and
maximum range, instantaneous position and velocity.
54
2.4 Projectile motion (2 hours)
θuux cos
θuuy sin
0xa
gay 

55. 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.
55
v
u

sx= R
sy=H
ux
v2y
uy
v1x
v1y
v2x
v1
1
v2
2
t1 t2
B
A
P Q
C
y
x
Figure 2.9
Simulation 2.5

56. PHYSICS CHAPTER 2
56
 From Figure 2.9,
 The x-component of velocity along AC (horizontal) at any
point is constant,
 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
θuux cos
θuuy sin

57. PHYSICS CHAPTER 2
Velocity Point P Point Q
x-comp.
y-comp.
magnitude
direction
57
 Table 2.2 shows the x and y-components, magnitude and
direction of velocities at points P and Q.
11 gtuv yy 
θuuv xx1 cos
22 gtuv yy 
θuuv xx2 cos
   2
y1
2
x11 vvv 






 
x1
y11
1
v
v
θ tan
   2
y2
2
x22 vvv 






 
x2
y21
2
v
v
θ tan
Table 2.2

58. PHYSICS CHAPTER 2
58
 The ball reaches the highest point at point B at velocity, v
where
 x-component of the velocity,
 y-component of the velocity,
 y-component of the displacement,
 Use
2.4.1 Maximum height, H
θuuvv xx cos
0yv
yyy gsuv 222

  gHu 2sin0
2
 
g
u
H
2
sin22


Hsy 

59. PHYSICS CHAPTER 2
59
 At maximum height, H
 Time, t = t’ and vy= 0
 Use
2.4.2 Time taken to reach maximum height, t’
gtuv yy 
  'sin0 tgu  
g
u
t
sin
'
2.4.3 Flight time, t (from point A to point C)
'2 tt 
g
θu
t
sin2


60. PHYSICS CHAPTER 2
tus xx 
60
 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
2.4.4 Horizontal range, R and value of R maximum
cosuvu xx 
  tuR  cos
  






g
u
uR


sin2
cos
  cossin2
2
g
u
R 
and Rsx 

61. PHYSICS CHAPTER 2
61
 From the trigonometry identity,
thus
 The value of R maximum when  = 45 and sin 2 = 1
therefore
 cossin22sin 
2sin
2
g
u
R 
g
u
R
2
max 
Simulation 2.6

62. PHYSICS CHAPTER 2
62
 Figure 2.10 shows a ball bearing rolling off the end of a table
with an initial velocity, u in the horizontal direction.
 Horizontal component along path AB.
 Vertical component along path AB.
2.4.5 Horizontal projectile
h
xA B
u u
v
xv
yv
Figure 2.10
constantvelocity,  xx vuu
xsx nt,displaceme
0uy velocity,initial
hsy nt,displaceme
Simulation 2.7

63. PHYSICS CHAPTER 2
63
Time taken for the ball to reach the floor (point B), t
 By using the equation of freely falling bodies,
Horizontal displacement, x
 Use condition below :
2
yy gttus
2
1

2
gt0h
2
1

g
h
t
2

The time taken for the
ball free fall to point A
The time taken for the
ball to reach point B
=
(Refer to Figure 2.11)
Figure 2.11

64. PHYSICS CHAPTER 2
64
 Since the x-component of velocity along AB is constant, thus
the horizontal displacement, x
 Note :
 In solving any calculation problem about projectile motion,
the air resistance is negligible.
tus xx 









g
h
ux
2
and xsx 

65. PHYSICS CHAPTER 2
65
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.
Example 2.9 :
Figure 2.12 xO
u
 = 60.0
y
R
H
v2y
v1x
v1y v2x
Q
v1
P
v2

66. PHYSICS CHAPTER 2
66
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
sm1000.60cos200 
 
xu
1
sm1730.60sin200 
 
yu

67. PHYSICS CHAPTER 2
67
Solution :
a. i. position of the ball when t = 2.0 s ,
Horizontal component :
Vertical component :
therefore the position of the ball is (200 m, 326 m)
2
yy gttus
2
1

tus xx 

68. PHYSICS CHAPTER 2
68
Solution :
a. ii. magnitude and direction of ball’s velocity at t = 2.0 s ,
Horizontal component :
Vertical component :
Magnitude,
Direction,
gtuv yy 
1
xx uv 
 sm100
from positive x-axis anticlockwise

69. PHYSICS CHAPTER 2
69
Solution :
b. i. At the maximum height, H :
Thus the time taken to reach maximum height is given by
ii. Apply
gtuv yy 
0yv
2
yy gttus
2
1


70. PHYSICS CHAPTER 2
70
Solution :
c. Flight time = 2(the time taken to reach the maximum height)
Hence the horizontal range, R is
d. When the ball reaches point P thus
The velocity of the ball at point P,
Horizontal component:
Vertical component:
s35.2t
 17.62t
tus xx 
1
1 sm100 
 xx uv
0ys
gtuv yy 1

71. PHYSICS CHAPTER 2
71
Solution :
Magnitude,
Direction,
therefore the direction of ball’s velocity is
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 :
from positive x-axis anticlockwise
tus xx 

72. PHYSICS CHAPTER 2
72
Solution :
Vertical component :
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 :
Vertical component :
2
yy gttus
2
1

gtuv yy 2
1
2 sm100 
 xx uv

73. PHYSICS CHAPTER 2
73
Solution :
Magnitude,
Direction,
therefore the direction of ball’s velocity is
from positive x-axis anticlockwise
2
2
2
22 yx vvv 






 
x
y
v
v
θ
2
21
tan

74. PHYSICS CHAPTER 2
74
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 :
Example 2.10 :
300 m
d
1
sm50 
u
X

75. PHYSICS CHAPTER 2
75
Solution :
The parcel’s velocity = plane’s velocity
thus
a. The vertical displacement is given by
Thus the flight time of the parcel is
1
sm50 
 uux
1
sm50 
u
m300ys
and
1
sm0 
yu
2
2
1
gttus yy 

76. PHYSICS CHAPTER 2
76
Solution :
b. The components of velocity of impact of the parcel :
Horizontal component :
Vertical component :
Magnitude,
Direction,
therefore the direction of parcel’s velocity is
1
sm50 
 xx uv
gtuv yy 
from positive x-axis anticlockwise

77. PHYSICS CHAPTER 2
77
Solution :
c. Let the distance from X to the point of impact is d.
Thus the distance, d is given by
tus xx 

78. PHYSICS CHAPTER 2
78
Figure 2.13
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.
ANS. : 10.7 m s1
Exercise 2.4 :

79. PHYSICS CHAPTER 2
79
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.
Exercise 2.4 :

80. PHYSICS CHAPTER 2
80
THE END…
Next Chapter…
CHAPTER 3 :
Momentum and Impulse