note chapter2 student

PHYSICS CHAPTER 2

1

xsr

ysr

xvr

yvr

xar

yar

gr

CHAPTER 2:CHAPTER 2:

Kinematics of linear motionKinematics of linear motion

(5 hours)(5 hours)

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sPHYSICS CHAPTER 2

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2.0 Kinematics of Linear motion

� is defined as the

� 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)

PHYSICS CHAPTER 2

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

�� DefineDefine distance, displacement, speed, velocity, distance, displacement, speed, velocity,

acceleration and related parameters: uniform velocity, acceleration and related parameters: uniform velocity,

average velocity, instantaneous velocity, uniform average velocity, instantaneous velocity, uniform

acceleration, average acceleration and instantaneous acceleration, average acceleration and instantaneous

acceleration.acceleration.

�� SketchSketch graphs of displacementgraphs of displacement--time, velocitytime, velocity--time and time and

accelerationacceleration--time.time.

Learning Outcome:

2.1 Linear Motion (1 hour)

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PHYSICS CHAPTER 2

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2.1. Linear motion (1-D)

2.1.1. Distance, d� scalar quantity.


� For example :

� The length of the path from P to Q is

P

Q

PHYSICS CHAPTER 2

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� vector quantity

� is defined as

� The S.I. unit of displacement is

� Example 1:

An object P moves 20 m to the east after that 10 m to the south

and finally moves 30 m to west. Determine the displacement of P

relative to the original position.

2.1.2 Displacement,sr

N

EW

S

O

PHYSICS CHAPTER 2

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The magnitude of the displacement is given by

and its direction is

2.1.3 Speed, v� is defined the

� scalar quantity.

� Equation:

=speed

PHYSICS CHAPTER 2

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� is a vector quantity.

� The S.I. unit for velocity is m s-1.

Average velocity, Average velocity, vvavav

� is defined as

� Equation:

� Its direction is in the of the change in of the change in

displacementdisplacement.

2.1.4 Velocity,vr

=avv

=avv

PHYSICS CHAPTER 2

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Instantaneous velocity, Instantaneous velocity, vv

� is defined as

� Equation:

� An object is moving in uniform velocitymoving in uniform velocity if

t

s

0tv

∆

∆

→∆=

limit

=dt

ds

PHYSICS CHAPTER 2

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� Therefore

Q

s

t0

s1

t1

The gradient of the tangent to the curve at point Q

=

PHYSICS CHAPTER 2

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� vector quantity

� The S.I. unit for acceleration is m s-2.

Average acceleration, Average acceleration, aaavav

� is defined as

� Equation:

� Its direction is in the

� The accelerationacceleration of an object is when the

of velocity changes at a of velocity changes at a

and along and along

2.1.5 Acceleration, ar

=ava

=ava

PHYSICS CHAPTER 2

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Instantaneous acceleration, Instantaneous acceleration, aa


� Equation:

� An object is moving in uniform acceleration moving in uniform acceleration if

t

v

0ta

∆

∆

→∆=

limit

=dt

dv

PHYSICS CHAPTER 2

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Deceleration,Deceleration, aa

� is a

� The object is meaning the

� Therefore

v

t

Q

0

v1

t1

The gradient of the tangent to the curve at point Q

=

PHYSICS CHAPTER 2

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Displacement against time graph (Displacement against time graph (ss--tt))

2.1.6 Graphical methods

s

t0

s

t0(a) Uniform velocity (b) The velocity increases with time

(c)

s

t0

PHYSICS CHAPTER 2

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Velocity versus time graph (Velocity versus time graph (vv--tt))

� The gradient at point A is positive –

� The gradient at point B is zero –

� The gradient at point C is negative –

t1 t2

v

t0(a) t2t1

v

t0(b)

t1 t2

v

t0(c)

PHYSICS CHAPTER 2

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� From the equation of instantaneous velocity,

Therefore

dt

dsv =

∫∫ = vdtds

∫=2

1

t

tvdts

Simulation 2.1 Simulation 2.2 Simulation 2.3

PHYSICS CHAPTER 2

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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.

Example 2 :

0 2 4 6 8 10 12 14 t (s)

2

4

6

8

10

s (cm)

Figure 2.1Figure 2.1

PHYSICS CHAPTER 2

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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-1) 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 3 :

05 10 15 20 25 30 35 t (s)

-4

-2

2

4

v (m s −1)


40 45 50

PHYSICS CHAPTER 2

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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 mANS. : 6 m

Exercise 2.1 :

PHYSICS CHAPTER 2

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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 sANS. : 0.4 m s−−−−−−−−22,0 m s,0 m s−−−−−−−−22,,--0.267 m s0.267 m s−−−−−−−−22, 80 m, 800 m, 120 m; , 80 m, 800 m, 120 m;

6.67 m s6.67 m s−−−−−−−−11..

Exercise 2.1 :

PHYSICS CHAPTER 2

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�� Derive and applyDerive and apply equations of motion with uniform equations of motion with uniform acceleration:acceleration:

Learning Outcome:

2.2 Uniformly accelerated motion (1 hour)

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atuv +=

2

2

1atuts +=

asuv 222+=

PHYSICS CHAPTER 2

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2.2. Uniformly accelerated motion

� From the definition of average acceleration, uniform (constantconstant) acceleration is given by

where v : final velocity

u : initial velocity

a : uniform (constant) acceleration

t : time

(1)

t

uva

−=

PHYSICS CHAPTER 2

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� From equation (1), the velocity-time graph is shown in figure

2.4:

� From the graph,

The displacement after time, s =

=

� Hence,

velocity

0

v

u

timetFigure 2.4Figure 2.4

(2)

PHYSICS CHAPTER 2

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

� From eq. (1),

� From eq. (2),

( )[ ]tatuus ++=2

1

(3)

( ) atuv =−

( )t

suv

2=+

multiply

( )( ) ( )att

suvuv

=−+

2

(4)

PHYSICS CHAPTER 2

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� Notes:

� equations (1) – (4) can be used if the motion in a straight motion in a straight

line with constant acceleration.line with constant acceleration.

� For a body moving at constant velocity, ( ( aa = 0)= 0) the

equations (1) and (4) become

Therefore the equations (2) and (3) can be written as

uv =

PHYSICS CHAPTER 2

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A plane on a runway takes 16.2 s over a distance of 1200 m to

take off from rest. Assuming constant acceleration during take off,

calculate

a. the speed on leaving the ground,

b. the acceleration during take off.

Solution :Solution :

Example 4 :

PHYSICS CHAPTER 2

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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.


Example 5 :

21 ms 2 0; ;constant s m 30 −−==== ccb auv

PHYSICS CHAPTER 2

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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 =2.00 s, determine

a. the displacement of the particle,

b. Its velocity, and

c. Its acceleration.

Example 6 :

ttts23 243 +−=

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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 sANS. : 4.53 s; 14.1 m s−−−−−−−−11

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 sANS. : 14.4 s

Exercise 2.2 :

PHYSICS CHAPTER 2

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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 sANS. : 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 mANS. : 1.73 m

Exercise 2.2 :

PHYSICS CHAPTER 2

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�� Describe and useDescribe and use equations for freely falling bodies.equations for freely falling bodies.

�� For For upward and downwardupward and downward motion, usemotion, use

aa = = −−−−−−−−gg = = −−−−−−−−9.81 m s9.81 m s−−−−−−−−22

Learning Outcome:

2.3 Freely falling bodies (1 hour)

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PHYSICS CHAPTER 2

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2.3. Freely falling bodies� is defined as

� In the earth’s gravitational field, the constant acceleration

� known as or or

� the value is

� the direction is

� Note:

� In solving any problem involves freely falling bodies or free

fall motion, the

PHYSICS CHAPTER 2

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� Sign convention:

� Table 2.1 shows the equations of linear motion and freely

falling bodies.

Table 2.1Table 2.1

Freely falling bodiesLinear motion

From the sign convention

thus,

PHYSICS CHAPTER 2

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� 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

when the ball reaches the

u

vFigure 2.5Figure 2.5

PHYSICS CHAPTER 2

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� The graphs in figure 2.6 show

the motion of the ball moves

up and down.

Derivation of equationsDerivation of equations

� At the maximum height or

displacement, H where t = t1,

its velocity,

hence

therefore the time taken for

the ball reaches H,


−u

t0

v

u

t1 2t1

a

t0

−g

t1 2t1

t

s

0

H

t1 2t1

v =0

gtuv −=

1gtu −=0

PHYSICS CHAPTER 2

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� To calculate the maximum height or displacement, H:

use either

maximum height,

� Another form of freely falling bodies expressions are

211 gtuts

2

1−=

gsuv22 2−=

Where s = H

gHu 20 2−=

OROR

PHYSICS CHAPTER 2

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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 stone when it reaches point D.

(Given g = 9.81 m s−2)

Example 7 :

A

B

C

D

u =10.0 m s−1

30.0 m


PHYSICS CHAPTER 2

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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)


Example 8 :

uy = 0 m s−1

150 mm 150−=ys

PHYSICS CHAPTER 2

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b. The book’s velocity is given by

Therefore the book’s velocity is

m 150−=ys

0=yu

?=yv

PHYSICS CHAPTER 2

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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 sANS. : 1.79 s; 25.6 m s−−−−−−−−11

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 mANS. : 1.75 m

Exercise 2.3 :

m 2.2


to travel this

distance took

0.30 s

PHYSICS CHAPTER 2

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�� Describe and useDescribe and use equations for projectile, equations for projectile,

�� CalculateCalculate: time of flight, maximum height, range and : time of flight, maximum height, range and maximum range, instantaneous position and velocity.maximum range, instantaneous position and velocity.

Learning Outcome:

2.4 Projectile motion (2 hours)

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s

θuux cos=

θuu y sin=

0=xa

gay −=

PHYSICS CHAPTER 2

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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.

t1 t2

B

A

P Q

C

y

xFigure 2.9Figure 2.9

PHYSICS CHAPTER 2

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� From figure 2.9,

� The xx--component of velocitycomponent of velocity along AC (horizontal) at any

point is constant,constant,

� The yy--component (vertical) of velocity variescomponent (vertical) of velocity varies from one

point to another point along AC.

but the y-component of the initial velocity is given by

PHYSICS CHAPTER 2

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� Table 2.2 shows the x and y-components, magnitude and

direction of velocities at points P and Q.

direction

magnitude

y-comp.

x-comp.

Point QPoint PVelocity

Table 2.2Table 2.2

PHYSICS CHAPTER 2

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� 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 cos===0=yv

yyy gsuv 222−=

( ) gHu 2sin02

−= θ

Hsy =

PHYSICS CHAPTER 2

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� 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 ∆−= θ

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

'2 tt ∆=∆

PHYSICS CHAPTER 2

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� 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

tus xx =

θcosuvu xx ==

and Rsx =

PHYSICS CHAPTER 2

47

� From the trigonometry identity,

thus

� The value of R maximum when θθθθθθθθ = = 4545°°°°°°°° and sin 2sin 2θθθθθθθθ = = 11therefore

θθθ cossin22sin =

PHYSICS CHAPTER 2

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� 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


=xu velocity,=xs nt,displaceme

=yu velocity,initial=ys nt,displaceme

PHYSICS CHAPTER 2

49

Time taken for the ball to reach the floor (point B), Time taken for the ball to reach the floor (point B), tt

� By using the equation of freely falling bodies,

Horizontal displacement, Horizontal displacement, xx

� Use condition below :

2yy gttus

2

1−=

2gt0h

2

1−=−

(Refer to figure 2.11)


PHYSICS CHAPTER 2

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� 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 negligibleair resistance is negligible.

tus xx = and xsx =

PHYSICS CHAPTER 2

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Figure 2.12 shows a ball thrown by superman

with an initial speed, u = 200 m s-1 and makes an

angle, θ = 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 9 :

Figure 2.12Figure 2.12 xO

u

θ = 60.0°

y

R

H

v2y

v1x

v1y v2xQ

v1

P

v2

PHYSICS CHAPTER 2

52

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)

PHYSICS CHAPTER 2

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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)


Example 10 :

300 m

d

1s m 50 −=u

X

PHYSICS CHAPTER 2

54


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 sANS. : 10.7 m s−−−−−−−−11

Exercise 2.4 :

PHYSICS CHAPTER 2

55

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 mANS. : 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.ANS. : 10.3 m below the original level.

Exercise 2.4 :

PHYSICS CHAPTER 2

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Force, Momentum and Impulse

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