kinematic of linear motion

PHYSICS CHAPTER 2

CHAPTER 2:CHAPTER 2:Kinematics of linear motionKinematics of linear motion

1

PHYSICS CHAPTER 2

Kinematics of linear motionKinematics of linear motion

2.1 Linear Motion2.2 Uniformly Accelerated Motion2.3 Free Falling Body2.4 Projectile Motion

2

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 and distinguish between and distinguish between

i.i. distance and displacement distance and displacement ii.ii. speed and velocityspeed and velocityiii.iii. instantaneous velocity, average velocity and uniform instantaneous velocity, average velocity and uniform

velocity.velocity.iv.iv. instantaneous acceleration, average acceleration and instantaneous acceleration, average acceleration and

uniform acceleration. uniform acceleration. SketchSketch graphs of displacement-time, velocity-time and graphs of displacement-time, velocity-time and

acceleration-time.acceleration-time. DetermineDetermine the distance travelled, displacement, velocity the distance travelled, displacement, velocity

and acceleration from appropriate graphs.and acceleration from appropriate graphs.

Learning Outcome:2.1 Linear Motion

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2.1. Linear motion (1-D)

2.1.1. Distance, d scalar quantity. is defined as the length of actual path between two pointslength of actual path between two points. For example :

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

P

Q

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vector quantity is defined as the distance between initial point and final the distance between initial point and final

point in a straight linepoint in a straight line. The S.I. unit of displacement is metre (m).

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.Solution :Solution :

2.1.2 Displacement,s

N

EW

S

O

P

θ

θ

20 m

10 m

10 m 20 m

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The magnitude of the displacement is given by

and its direction is

2.1.3 Speed, v is defined the rate of change of distancerate of change of distance. scalar quantity. Equation:

interval timedistance of changespeed =

ΔtΔdv =

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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 the rate of change of displacementthe rate of change of displacement. Equation:

Its direction is in the same direction of the change in same direction of the change in displacementdisplacement.

2.1.4 Velocity,v

interval timentdisplaceme of change=avv

ΔtΔsvav =

12

12av tt

ssv−−=

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Instantaneous velocity, Instantaneous velocity, vv is defined as the instantaneous rate of change of the instantaneous rate of change of

displacementdisplacement. Equation:

An object is moving in uniform velocitymoving in uniform velocity if

ts

0tv

∆∆

→∆=

limit

constant=dtds

dtdsv =

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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

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

Average acceleration, Average acceleration, aaavav

is defined as the rate of change of velocitythe rate of change of velocity. Equation:

Its direction is in the same direction of motionsame direction of motion. The accelerationacceleration of an object is uniformuniform when the magnitude magnitude

of velocity changes at a constant rate and along fixed of velocity changes at a constant rate and along fixed direction.direction.

2.1.5 Acceleration, a

interval time velocityof change=ava

12

12av tt

vva−−=

ΔtΔvaav =

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Instantaneous acceleration, Instantaneous acceleration, aa is defined as the instantaneous rate of change of velocityinstantaneous rate of change of velocity. Equation:

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

tv

0ta

∆∆

→∆=

limit

constant=dtdv

2

2

dtsd

dtdva ==

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Deceleration,Deceleration, aa is a negative accelerationnegative acceleration. The object is slowing downslowing down meaning the speed of the object speed of the object

decreases with timedecreases 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

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Displacement against time graph (Displacement against time graph (s-ts-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.

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Velocity versus time graph (Velocity versus time graph (v-tv-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 velocityUniform acceleration

Area under the v-t graph = displacement

BC

A

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From the equation of instantaneous velocity,

Therefore

dtdsv =

∫∫ = vdtds

∫= 2

1

t

tvdts

graph under the area dedsha tvs −=

Simulation 2.1 Simulation 2.2 Simulation 2.3

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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

68

10

s (cm)

Figure 2.1Figure 2.1

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Solution :Solution :0 to 6 s :

6 to 10 s : 10 to 14 s :

b.

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

0.68

1.50

v (cm s−1)

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Solution :Solution :c.

d.

12

12

ttssvav −

−=

s 14 tos 10 from velocity average=v

12

12

ttssv

−−=

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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

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Solution :Solution :a. 0 to 5 s : Lift moves upward from rest with

acceleration of 0.4 m s−2. 5 to 15 s : The velocity of the lift from 2 m s−1 to

4 m s−1 but the acceleration to 0.2 m s−2. 15 to 20 s : Lift 20 to 25 s : Lift 25 to 30 s : Lift 30 to 35 s : Lift moves

35 to 40 s : Lift moving

40 to 50 s :

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Solution :Solution :b.

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

-0.4-0.2

0.2

0.6

a (m s−2)

-0.6

-0.8

0.8

0.4

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Solution :Solution :c. i.

05 10 15 20 25 30 35 t (s)

-4-2

2

4

v (m s −1)

40 45 50A1

A2 A3

A4 A5

v-t ofgraph under the area distance Total =54321 AAAAA ++++=

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )45152145

214105

211042

2152

21distance Total +++++++=

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Solution :Solution :c. ii.

d.

v-t ofgraph under the areant Displaceme =

54321 AAAAA ++++=

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )45152145

214105

211042

2152

21ntDisplaceme −++−+++++=

12

12

ttvvaav −

−=

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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 :

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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 s,-0.267 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 :

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

acceleration:acceleration:

Learning Outcome:2.2 Uniformly accelerated motion

atuv +=2

21 atuts +=

asuv 222 +=

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2.2. Uniformly accelerated motion From the definition of average acceleration, uniform (constantconstant)

acceleration is given by

where v : final velocityu : initial velocity

a : uniform (constant) accelerationt : time

atuv += (1)

tuva −=

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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 = shaded area under the

graph = the area of trapezium

Hence,

velocity

0

v

utimetFigure 2.4Figure 2.4

( )tvu21s += (2)

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

From eq. (1),

From eq. (2),

( )[ ]tatuus ++=21

(3)2

21 atuts +=

( ) atuv =−

( )tsuv 2=+

multiply

( ) ( ) ( )attsuvuv

=−+ 2

asuv 222 += (4)

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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 asuv =

vts = constant velocityconstant velocity

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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, calculatea. the speed on leaving the ground,b. the acceleration during take off.Solution :Solution :

a. Use

Example 4 :

s 2.16=t

?=v

( )tvus +=21

0=u

m 1200=s

?=a

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Solution :Solution :b. By using the equation of linear motion,

asuv 222 +=

OROR2

21 atuts +=

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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. Determinea. 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 :Solution :

a. Given Use

Example 5 :

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

cccc tauv +=1s m 30 −== bc vv

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b.

From the diagram,

c

b

1s m 30 −=bv

0=cus 0=bt s 5=bt

2s m 2 −=cab

bvb

c

bv

ttb =bc ss =

bc ss =

bbcccc tvtatu =+ 2

21

Thereforetvs bc =

;ttb = 5−= ttc

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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, determinea. the displacement of the particle,b. Its velocity, andc. Its acceleration.Solution :Solution :a. t =2.00 s ;

Example 6 :

ttts 23 243 +−=

ttts 23 243 +−=

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Solution :Solution :b. Instantaneous velocity at t = 2.00 s,

Use

Thus

dtdsv =

( )tttdtdv 243 23 +−=

( ) ( ) 22.0082.009 2 +−=v

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Solution :Solution :c. Instantaneous acceleration at t = 2.00 s,

Use

Hence

dtdva =

( ) 82.0018 −=a

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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 :

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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 s4. 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 :

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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: Describe and useDescribe and use equations for free falling body. equations for free falling body.

For For upward and downwardupward and downward motion, use motion, useaa = = −−gg = = −−9.81 m s9.81 m s−−22

Learning Outcome:2.3 Free falling body

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2.3. Free falling body is defined as the vertical motion of a body at constant the vertical motion of a body at constant

acceleration, acceleration, gg under gravitational field under gravitational field without air without air resistanceresistance..

In the earth’s gravitational field, the constant acceleration known as acceleration due to gravityacceleration due to gravity or free-fall free-fall

accelerationacceleration or gravitational accelerationgravitational acceleration. the value is gg = 9.81 m s= 9.81 m s−−22

the direction is towards the centre of the earth towards the centre of the earth (downward).(downward).

Note: In solving any problem involves freely falling bodies or free

fall motion, the assumption made is ignore the air assumption made is ignore the air resistanceresistance.

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Sign convention:

Table 2.1 shows the equations of linear motion and freely falling bodies.

Table 2.1Table 2.1

Linear motion Freely falling bodiesatuv += gtuv −=

as2uv 22 += gs2uv 22 −=2at

21uts += 2gt

21uts −=

+

- +

-

From the sign convention thus,

ga −=

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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 velocity velocity decreases to zerodecreases to zero when the ball reaches the maximum maximum height, height, HH.

H

uv

velocity = 0


uv =

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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,


t0

vu

−u

t1 2t1

t0

a

−g

t1 2t1

t

s

0

H

t1 2t1

v =0

gtuv −=1gtu −=0

0=v

gut1 =

Simulation 2.4

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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

21−=

gsuv 22 2−=Where s = H

gHu 20 2 −=

OROR

guH2

2

=

gtuv −=gsuv 222 −=

2

21 gtuts −=

gtuv yy −=yyy gsuv 222 −=2

21 gttus yy −=

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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. Calculatea. 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


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Solution :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

y2y

2y gsuv 2−=

2yy gttus

21−=

A

B

C

D

u

30.0 m

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

By using

2yy gttus

21−=

A

B

C

D

u

30.0 m2a

4acbb 2 −±−=t

ORTime don’t Time don’t have have negative negative value.value.

a b c

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Solution :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

gtuv yy −=( ) ( ) ( )3.699.8110.0 −=yv

OR

y2

y2

y gsuv 2−=( ) ( )( )30.09.81210.0 −−= 22

yv

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A book is dropped 150 m from the ground. Determinea. 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 :Solution :

a. The vertical displacement issy = −150 m

Hence

Example 8 :

uy = 0 m s−1

150 mm 150−=ys

2yy gttus

21−=

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Solution :Solution :b. The book’s velocity is given by

Therefore the book’s velocity is

gtuv yy −=

OR

y2

y2

y gsuv 2−=m 150−=ys

0=yu

?=yv

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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

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

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

Learning Outcome:2.4 Projectile motion

θuux cos=θuu y sin=

0=xagay −=

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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.

v

u

θsx= R

sy=H

ux

v2yuy

v1x

v1y

v2x

v1

θ1

v2

θ2

t1 t2

B

A

P Q

C

y

xFigure 2.9Figure 2.9

Simulation 2.5

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From figure 2.9, The x-component of velocityx-component of velocity along AC (horizontal) at any

point is constant,constant,

The y-component (vertical) of velocity variesy-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=

θuu y sin=

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Table 2.2 shows the x and y-components, magnitude and direction of velocities at points P and Q.

Velocity Point P Point Q

x-comp.

y-comp.

magnitude

direction

11 gtuv yy −=θuuv xx1 cos==

22 gtuv yy −=θuuv xx2 cos==

( ) ( )2y1

2x11 vvv +=

= −

x1

y111 v

vθ tan

( ) ( )2y2

2x22 vvv +=

= −

x2

y212 v

vθ tan

Table 2.2Table 2.2

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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 −= θ

guH

2sin 22 θ=

Hsy =

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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

gut θsin' =∆

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

'2 tt ∆=∆

gθut sin2=∆

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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 ==

( ) ( )tuR ∆= θcos

( )

=

guuR θθ sin2cos

( )θθ cossin22

guR =

and Rsx =

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60

From the trigonometry identity,

thus

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

θθθ cossin22sin =

θ2sin2

guR =

guR

2

max =

Simulation 2.6

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61

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

vxv

yv


constant velocity, === xx vuuxsx = nt,displaceme

0u y = velocity,initialhsy −= nt,displaceme

Simulation 2.7

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62

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

21−=

2gt0h21−=−

ght 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)


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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 =

=

ghux 2

and xsx =

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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. Determinea. 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 v2xQv1

P

v2

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65

b. the time taken for the ball reaches the maximum height, H and

calculate the value of H.

c. the horizontal range, Rd. 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 :Solution :The component of Initial velocity :

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66

Solution :Solution :a. i. position of the ball when t = 2.0 s ,

Horizontal component :

Vertical component :

therefore the position of the ball is

2yy gttus

21−=

tus xx =

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Solution :Solution :a. ii. magnitude and direction of ball’s velocity at t = 2.0 s ,



Magnitude,

Direction,

gtuv yy −=

1xx uv −== s m 100

( ) ( ) 2222 153100 +=+= yx vvv

=

= −−

100153tantan 11

x

y

vv

θ

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68

Solution :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

gttus yy 21−=

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Solution :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 thusThe velocity of the ball at point P,Horizontal component:Vertical component:

( )17.62=t

tus xx =

11 s m 100 −== xx uv

0=ys

gtuv yy −=1

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Solution :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,


( ) ( ) 2221

211 172100 −+=+= yx vvv

−=

= −−

100172tantan 1

1

11

x

y

vv

θ

300=θ from positive x-axis anticlockwisefrom positive x-axis anticlockwise

tus xx =

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71

Solution :Solution :Vertical component :

therefore the position of the ball is (4500 m, (4500 m, −−2148 m)2148 m)e. ii. magnitude and direction of ball’s velocity at t = 45.0 s ,



2yy gttus

21−=

gtuv yy −=2

12 s m 100 −== xx uv

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72

Solution :Solution :Magnitude,

Direction,

therefore the direction of ball’s velocity is

( ) ( ) 222 269100 −+=v

22

222 yx vvv +=

= −

x

y

vv

θ2

21tan

PHYSICS CHAPTER 2

73

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. Calculatea. 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 :Solution :

Example 10 :

300 m

d

1s m 50 −=u

X

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74

Solution :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 −=uand 1s m 0 −=yu

2

21 gttus yy −=

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75

Solution :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 uv

( ) ( )7.829.810 −=yvgtuv yy −=

−=

= −−

506.77tantan 11

x

y

vv

θ

( ) ( ) 2222 6.7750 −+=+= yx vvv

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76

Solution :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

77


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

78

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. Calculatea. 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

79

THE END…Next Chapter…

CHAPTER 3 :Force, Momentum and Impulse

kinematic of linear motion

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