PHYSICS CHAPTER 2

1

CHAPTER 2:

Kinematics of Linear Motion

(5 hours)

xs

ys

xv

yv

xa

ya

g

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 :

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)

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

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

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:

interval time

distance of changespeed

6

Δt

Δdv

m 181015 22 OP

south west tofrom 5610

15tan 1

θ

PHYSICS CHAPTER 2

v

7

interval time

ntdisplaceme of changeavv

12

12av

tt

ssv

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

Δsvav

PHYSICS CHAPTER 2

constantdt

ds

8

t

s

0tv

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

dsv

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

PHYSICS CHAPTER 2

a

10

interval time

velocityof changeava

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

12av

tt

vva

Δt

Δvaav

PHYSICS CHAPTER 2

constantdt

dv

11

t

v

0ta

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

dva

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

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

RP

The direction of

velocity is changing.

Gradient at point R is negative.

Gradient at point Q is zero.

The velocity is zero.

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

BC

A

PHYSICS CHAPTER 2

dt

dsv

15

From the equation of instantaneous velocity,

Therefore

vdtds

2

1

t

tvdts

graph under the area dedsha tvs

Simulation 2.1 Simulation 2.2 Simulation 2.3

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

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 s1)

PHYSICS CHAPTER 2

18

Solution :

c.

d.

e. The distance travelled by the toy train is 10 cm.

12

12

tt

ssvav

s 14 tos 10 from velocity averagev

12

12

tt

ssv

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 s2) 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 :

05 10 15 20 25 30 35 t (s)

-4

-2

2

4

v (m s 1)

Figure 2.2

40 45 50

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 s1 to

4 m s1 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.

PHYSICS CHAPTER 2

21

Solution :

b.

t (s)5 10 15 20 25 30 35 40 45 500

-0.4

-0.2

0.2

0.6

a (m s2)

-0.6

-0.8

0.8

0.4

PHYSICS CHAPTER 2

22

Solution :

c. i.

05 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-t ofgraph under the area distance Total

54321 AAAAA

PHYSICS CHAPTER 2

23

Solution :

c. ii.

d.

v-t ofgraph under the areant Displaceme

54321 AAAAA

12

12

tt

vvaav

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 :

PHYSICS CHAPTER 2

25

2. A train pulls out of a station and accelerates steadily for 20 s

until its velocity reaches 8 m s1. 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 s2,0 m s2,-0.267 m s2, 80 m, 800 m, 120 m;

6.67 m s1.

Exercise 2.1 :

PHYSICS CHAPTER 2

Learning Outcome :

At the end of this chapter, students should be able to:

Derive and apply equations of motion with uniformacceleration:

26

2.2 Uniformly accelerated motion (1 hour)

atuv 2

2

1atuts

asuv 222

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

uva

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

timetFigure 2.4

tvu2

1s (2)

PHYSICS CHAPTER 2

29

By substituting eq. (1) into eq. (2) thus

From eq. (1),

From eq. (2),

tatuus 2

1

(3)2

2

1atuts

atuv

t

suv

2

multiply

att

suvuv

2

asuv 222 (4)

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

PHYSICS CHAPTER 2

31

A plane on a runway accelerates from rest and must attain takeoff

speed of 148 m s1 before reaching the end of the runway. The

plane’s acceleration is uniform along the runway and of value

914 cm s2. 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 :

?tasuv 222

0u

?s

2s m 14.9 a1s m 148 v

PHYSICS CHAPTER 2

32

Solution :

b. By using the equation of linear motion,

atuv

OR

2

2

1atuts

PHYSICS CHAPTER 2

33

A bus travelling steadily at 30 m s1 along a straight road passes a

stationary car which, 5 s later, begins to move with a uniform

acceleration of 2 m s2 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 ms 2 0; ;constant s m 30 ccb auv

cccc tauv

1s m 30 bc vv

PHYSICS CHAPTER 2

34

b.

From the diagram,

c

b

1s m 30 bv

0cu

s 0bt s 5bt

2s m 2 ca

bbv

b

c

bv

ttb

bc ss

bc ss

bbcccc tvtatu 2

2

1

Therefore

tvs bc ;ttb 5 ttc

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

PHYSICS CHAPTER 2

36

Solution :

b. Instantaneous velocity at t = 3 s,

Use

Thus

dt

dsv

322 ttdt

dv

PHYSICS CHAPTER 2

37

Solution :

c. Instantaneous acceleration at t = 3 s,

Use

Hence

dt

dva

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 s1

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 :

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 :

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 s2

40

2.3 Freely falling bodies (1 hour)

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 s2

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

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

2at2

1uts

2gt2

1uts

+

- +

-

From the sign convention

thus,

ga

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

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

0v

g

ut1

Simulation 2.4

PHYSICS CHAPTER 2

211 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

uH

2

2

gtuv

gsuv 222 2

2

1gtuts

gtuv yy

yyy gsuv 222 2

2

1gttus yy

PHYSICS CHAPTER 2

46

A ball is thrown from the top of a building is given an initial velocity

of 10.0 m s1 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 s2)

Example 2.7 :

A

B

C

D

u =10.0 m s1

30.0 m

Figure 2.7

PHYSICS CHAPTER 2

y2y

2y gsuv 2

H9.81210.002

47

Solution :

a. At the maximum height, H, vy = 0 and u = uy = 10.0 m s1 thus

b. From point A to C, the vertical displacement, sy= 0 m thus

m 5.10H

2yy gttus

2

1

A

B

C

D

u

30.0 m

PHYSICS CHAPTER 2

48

Solution :

c. From point A to D, the vertical displacement, sy= 30.0 m thus

By using

2yy gttus

2

1

s 3.69t

2tt 9.812

110.030.0

A

B

C

D

u

30.0 m

030.010.04.91 tt 2

OR s 1.66Time don’t

have

negative

value.

a b c

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

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 s1

150 mm 150ys

2yy gttus

2

1

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 2m 150ys

0yu

?yv

PHYSICS CHAPTER 2

52

1. A ball is thrown directly downward, with an initial speed of

8.00 m s1, 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 s1

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 :

m 2.2

Figure 2.8

to travel this

distance took

0.30 s

PHYSICS CHAPTER 2

53

1. A ball is thrown directly downward, with an initial speed of

8.00 m s1, 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 s1

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 :

m 2.2

Figure 2.8

to travel this

distance took

0.30 s

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 andmaximum range, instantaneous position and velocity.

54

2.4 Projectile motion (2 hours)

θuux cos

θuuy sin0xa

gay

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

xFigure 2.9

Simulation 2.5

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

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

2y1

2

x11 vvv

x1

y111

v

vθ tan

2y2

2

x22 vvv

x2

y212

v

vθ tan

Table 2.2

PHYSICS CHAPTER 2

58

The ball reaches the highest point at point B at velocity, vwhere

x-component of the velocity,

y-component of the velocity,

y-component of the displacement,

Use

2.4.1 Maximum height, H

θuuvv xx cos0yv

yyy gsuv 222

gHu 2sin02

g

uH

2

sin 22

Hs y

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

ut

sin'

2.4.3 Flight time, t (from point A to point C)

'2 tt

g

θut

sin2

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

uuR

sin2cos

cossin22

g

uR

and Rsx

PHYSICS CHAPTER 2

61

From the trigonometry identity,

thus

The value of R maximum when = 45 and sin 2 = 1therefore

cossin22sin

2sin2

g

uR

g

uR

2

max

Simulation 2.6

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

constant velocity, xx vuu

xsx nt,displaceme

0uy velocity,initialhsy nt,displaceme

Simulation 2.7

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 :

2yy gttus

2

1

2gt0h2

1

g

ht

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

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

hux

2

and xsx

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 v2xQ

v1

P

v2

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 :1s m 1000.60cos200

xu1s m 1730.60sin200

yu

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)

2yy gttus

2

1

tus xx

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

1xx uv s m 100

from positive x-axis anticlockwise

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

0yv

2yy gttus

2

1

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:

s 35.2t 17.62t

tus xx

1

1 s m 100 xx uv

0ys

gtuv yy 1

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

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 :

2yy gttus

2

1

gtuv yy 2

1

2 s m 100 xx uv

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

vθ

2

21tan

PHYSICS CHAPTER 2

74

A transport plane travelling at a constant velocity of 50 m s1 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

1s m 50 u

X

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

1s m 50 uux

1s m 50 u

m 300ys

and1s m 0 yu

2

2

1gttus yy

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

1s m 50 xx uvgtuv yy

from positive x-axis anticlockwise

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

PHYSICS CHAPTER 2

78

Figure 2.13

Use gravitational acceleration, g = 9.81 m s2

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 s1

Exercise 2.4 :

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 s1. 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 s1

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 :

PHYSICS CHAPTER 2

80

THE END…

Next Chapter…CHAPTER 3 :

Momentum and Impulse