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PHYSICS CHAPTER 2 1 CHAPTER 2: Kinematics of Linear Motion x s y s x v y v x a y a

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  • PHYSICS CHAPTER 2

    1

    CHAPTER 2:Kinematics of Linear Motion

    xs

    ys

    xv

    yv

    xa

    ya

  • 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

    Linear Motion

    Distance Displacement Speed Velocity Instantaneous acceleration Average acceleration

    3

  • PHYSICS CHAPTER 22.1. 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 2s

    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 2The 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 timedistance of changespeed =

    6

    t

    dv =

    m 181015 22 =+=OP

    south west tofrom 561015tan 1 =

    =

  • PHYSICS CHAPTER 2v

    7

    interval timentdisplaceme of change

    =avv

    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,

    t

    svav =

  • PHYSICS CHAPTER 2

    constant=dtds

    8

    ts

    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.

    dtdsv =

  • 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 2a

    10

    interval time velocityof change

    =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

    12av tt

    vva

    =

    t

    vaav =

  • PHYSICS CHAPTER 2

    constant=dtdv

    11

    tv

    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

    dtsd

    dtdva ==

  • 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 velocityUniform acceleration

    Area under the v-t graph = displacement

    BC

    A

  • PHYSICS CHAPTER 2

    dtdsv =

    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

    68

    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 ofb.

    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

    ttssvav

    =

    s 14 tos 10 from velocity average=v

    12

    12

    ttssv

    =

  • 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 of20 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

    of40 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 50A1

    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

    ttvvaav

    =

  • 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 2Learning 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

    21 atuts +=

    asuv 222 +=

  • PHYSICS CHAPTER 22.2. Uniformly accelerated motion From the definition of average acceleration, uniform (constant)

    acceleration is given by

    where v : final velocityu : initial velocitya : uniform (constant) accelerationt : 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

    utimetFigure 2.4

    ( )tvu21s += (2)

  • PHYSICS CHAPTER 2

    29

    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)

  • 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 asuv =

    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 planes acceleration is uniform along the runway and of value 914 cm s2. Calculatea. 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 +=

    0=u

    ?=s

    2s m 14.9 =a1s m 148 =v

  • PHYSICS CHAPTER 2

    32

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

    atuv +=

    OR2

    21 atuts +=

  • 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. 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 :

    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

    b1s m 30 =bv

    0=cus 0=bt s 5=bt

    2s m 2 =cab b

    vb

    c

    bv

    ttb =bc ss =

    bc ss =

    bbcccc tvtatu =+2

    21

    Thereforetvs 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, determinea. the displacement of the particle,b. Its velocity, andc. 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

    dtdsv =

    ( )322 ++= ttdtdv

  • PHYSICS CHAPTER 2

    37

    Solution :c. Instantaneous acceleration at t = 3 s,

    Use

    Hence

    dtdva =

  • 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 s12. An unmarked police car travelling a constant 95 km h-1 is

    passed by a speeder traveling 140 km h-1. Precisely 1.00 safter the speeder passes, the policemen steps on theaccelerator; if the police cars acceleration is 2.00 m s-2, howmuch time passes before the police car overtakes thespeeder (assumed moving at constant speed)?No. 44, pg. 41,Physics for scientists and engineers withmodern 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 withmodern physics, Douglas C. Giancoli,3rd edition.

    ANS. : 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 drivers 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 2Learning 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, usea = g = 9.81 m s2

    40

    2.3 Freely falling bodies (1 hour)

  • PHYSICS CHAPTER 22.3 Freely falling bodies is defined as the vertical motion of a body at constant

    acceleration, g under gravitational field without air resistance.

    In the earths 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 bodiesatuv += gtuv =

    as2uv 22 += gs2uv 22 =2at

    21uts += 2gt

    21uts =

    +

    - +

    -

    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

    uv

    velocity = 0

    Figure 2.5

    uv =

  • PHYSICS CHAPTER 2

    gtuv =

    44

    The graphs in Figure 2.6 showthe motion of the ball movesup 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

    vu

    u

    t1 2t1

    t0

    a

    g

    t1 2t1

    t

    s

    0

    H

    t1 2t1

    v =0

    1gtu =0

    0=v

    gut1 =

    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

    guH2

    2

    =

    gtuv =gsuv 222 =

    2

    21 gtuts =

    gtuv yy =yyy gsuv 2

    22 =2

    21 gttus 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. 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 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=

    ( ) ( )H9.81210.00 2 =

    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 5.10=H

    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 3.69=t

    ( ) ( ) 2tt 9.812110.030.0 =

    A

    B

    C

    D

    u

    30.0 m

    030.010.04.91 = tt 2

    OR s 1.66Time dont 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 balls velocity at D is

    A

    B

    C

    D

    u

    30.0 m

    OR

    y2

    y2

    y gsuv 2=

  • PHYSICS CHAPTER 2

    50

    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 :

    a. The vertical displacement is

    sy = 150 mHence

    Example 2.8 :

    uy = 0 m s1

    150 mm 150=ys

    2yy gttus 2

    1=

  • PHYSICS CHAPTER 2

    gtuv yy =

    51

    Solution :b. The books velocity is given by

    Therefore the books velocity is

    OR

    y2

    y2

    y gsuv 2=m 150=ys

    0=yu

    ?=yv

  • PHYSICS CHAPTER 2

    52

    1. A ball is thrown directly downward, with an initial speed of8.00 m s1, from a height of 30.0 m. Calculatea. the time taken for the ball to strike the ground,b. the balls speed when it reaches the ground.

    ANS. : 1.79 s; 25.6 m s12. 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 of8.00 m s1, from a height of 30.0 m. Calculatea. the time taken for the ball to strike the ground,b. the balls speed when it reaches the ground.

    ANS. : 1.79 s; 25.6 m s12. 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 2Learning 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)

    uux cos=uuy sin=

    0=xagay =

  • PHYSICS CHAPTER 22.4. 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

    v2yuy

    v1x

    v1y

    v2x

    v11

    v22

    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

    uux cos=

    uuy 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 =uuv xx1 cos==

    22 gtuv yy =uuv xx2 cos==

    ( ) ( )2y12x11 vvv +=

    =

    x1

    y111 v

    v tan

    ( ) ( )2y22x22 vvv +=

    =

    x2

    y212 v

    v 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

    uuvv xx cos===0=yv

    yyy gsuv 222 =

    ( ) gHu 2sin0 2 =

    guH

    2sin22

    =

    Hsy =

  • 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 =

    gut sin'=

    2.4.3 Flight time, t (from point A to point C)'2 tt =

    gut 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( )

    =

    guuR sin2cos

    ( ) cossin22

    guR =

    and Rsx =

  • PHYSICS CHAPTER 2

    61

    From the trigonometry identity,

    thus

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

    cossin22sin =

    2sin2

    guR =

    guR

    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

    vxv

    yv

    Figure 2.10

    constant velocity, === xx vuuxsx = 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=

    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)

    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 =

    =

    ghux 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. Determinea. 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 v2xQv1

    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, 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 :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 balls 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 =

    0=yv

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

    s 35.2=t( )17.62=t

    tus xx =

    11 s m 100

    == xx uv

    0=ys

    gtuv yy =1

  • PHYSICS CHAPTER 2

    71

    Solution :Magnitude,

    Direction,

    therefore the direction of balls 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 balls velocity at t = 45.0 s ,

    Horizontal component :

    Vertical component :

    2yy gttus 2

    1=

    gtuv yy =2

    12 s m 100

    == xx uv

  • PHYSICS CHAPTER 2

    73

    Solution :Magnitude,

    Direction,

    therefore the direction of balls velocity is

    from positive x-axis anticlockwise

    22

    222 yx vvv +=

    =

    x

    y

    vv

    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. 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 :

    Example 2.10 :

    300 m

    d

    1s m 50 =u

    X

  • PHYSICS CHAPTER 2

    75

    Solution :The parcels velocity = planes 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 300=ys

    and 1s m 0 =yu

    2

    21 gttus 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 parcels 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 s21. 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. 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 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 s1at 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 ENDNext ChapterCHAPTER 3 :

    Momentum and Impulse

    Slide Number 12.0 Kinematics of Linear motionLinear Motion2.1. Linear motion (1-D)Slide Number 5Slide Number 6Slide Number 7Slide Number 8Slide Number 9Slide Number 10Slide Number 11Slide Number 12Slide Number 13Slide Number 14Slide Number 15Slide Number 16Slide Number 17Slide Number 18Slide Number 19Slide Number 20Slide Number 21Slide Number 22Slide Number 23Slide Number 24Slide Number 25Learning Outcome :2.2. Uniformly accelerated motionSlide Number 28Slide Number 29Slide Number 30Slide Number 31Slide Number 32Slide Number 33Slide Number 34Slide Number 35Slide Number 36Slide Number 37Slide Number 38Slide Number 39Learning Outcome :2.3 Freely falling bodiesSlide Number 42Slide Number 43Slide Number 44Slide Number 45Slide Number 46Slide Number 47Slide Number 48Slide Number 49Slide Number 50Slide Number 51Slide Number 52Slide Number 53Learning Outcomes :2.4. Projectile motionSlide Number 56Slide Number 57Slide Number 58Slide Number 59Slide Number 60Slide Number 61Slide Number 62Slide Number 63Slide Number 64Slide Number 65Slide Number 66Slide Number 67Slide Number 68Slide Number 69Slide Number 70Slide Number 71Slide Number 72Slide Number 73Slide Number 74Slide Number 75Slide Number 76Slide Number 77Slide Number 78Slide Number 79Slide Number 80