CHAPTER 2: Kinematics of Linear Motion (5 hours)

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CHAPTER 2Kinematics of Linear Motion(5 hours)
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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)

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Learning Outcomes
2.1 Linear Motion (1 hour)

• 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.

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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 points.
• For example
• The length of the path from P to Q is 25 cm.

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2.1.2 Displacement,

• 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

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• 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

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2.1.4 Velocity,

• 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.

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• 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.

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• Therefore

The gradient of the tangent to the curve at point
Q the instantaneous velocity at time, t t1
Q
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2.1.5 Acceleration,

• 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.

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• 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.

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• Deceleration, a
• is a negative acceleration.
• The object is slowing down meaning the speed of
  the object decreases with time.
• Therefore

The gradient of the tangent to the curve at point
Q the instantaneous acceleration at time, t
t1
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2.1.6 Graphical methods

• Displacement against time graph (s-t)

Gradient increases with time
Gradient constant
(a) Uniform velocity
(b) The velocity increases with time
(c)
Gradient at point R is negative.
The direction of velocity is changing.
Gradient at point Q is zero.
The velocity is zero.
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• Velocity versus time graph (v-t)
• The gradient at point A is positive a gt
  0(speeding up)
• The gradient at point B is zero a 0
• The gradient at point C is negative a lt
  0(slowing down)

Uniform acceleration
Uniform velocity
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• From the equation of instantaneous velocity,
• Therefore

Simulation 2.1
Simulation 2.2
Simulation 2.3
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Example 2.2
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.
Figure 2.1
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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.
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Solution c. d. e. The distance
travelled by the toy train is 10 cm.
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Example 2.3
A velocity-time (v-t) graph in Figure 2.2 shows
the motion of a lift. a. Describe
qualitatively the motion of the lift. b. Sketch a
graph of acceleration (m s?2) against time
(s). c. Determine the total distance travelled by
the lift and its displacement. d.
Calculate the average acceleration between 20 s
to 40 s.
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Solution a. 0 to 5 s Lift moves upward
from rest with a constant acceleration of
5 to 15 s The velocity of the lift
increases from 2 m s?1 to 4 m s?1 but the
acceleration decreasing to 15 to 20 s
Lift moving with constant velocity of 20 to
25 s Lift decelerates at a constant rate of
25 to 30 s Lift at rest or stationary.
30 to 35 s Lift moves downward with a
constant acceleration of 35 to 40 s
Lift moving downward with constant velocity
of 40 to 50 s Lift decelerates at a
constant rate of and comes to
rest.
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Solution b.
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Solution c. i.
A3
A2
A1
A4
A5
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Solution c. ii. d.
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Exercise 2.1

• 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

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Exercise 2.1

• A train pulls out of a station and accelerates
  steadily for 20 s until its velocity reaches 8 m
  s?1. It then travels at a constant velocity for
  100 s, then it decelerates steadily to rest in a
  further time of 30 s.
• a. Sketch a velocity-time graph for the
  journey.
• b. Calculate the acceleration and the distance
  travelled in each part of the journey.
• c. Calculate the average velocity for the
  journey.
• Physics For Advanced Level, 4th edition, Jim
  Breithaupt, Nelson Thornes, pg.15, no. 1.11
• ANS. 0.4 m s?2,0 m s?2,-0.267 m s?2, 80 m, 800
  m, 120 m
• 6.67 m s?1.

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Learning Outcome
2.2 Uniformly accelerated motion (1 hour)

• At the end of this chapter, students should be
  able to
• Derive and apply equations of motion with uniform
  acceleration

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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

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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,

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• By substituting eq. (1) into eq. (2) thus
• From eq. (1),
• From eq. (2),

multiply
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• 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

constant velocity
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Example 2.4
A plane on a runway accelerates from rest and
must attain takeoff speed of 148 m s?1 before
reaching the end of the runway. The planes
acceleration is uniform along the runway and of
value 914 cm s?2. Calculate a. the minimum length
of the runway required by the plane to
takeoff. b. the time taken for the plane cover
the length in (a). Solution a. Use
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Solution b. By using the equation of linear
motion,
OR
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Example 2.5
A bus travelling steadily at 30 m s?1 along a
straight road passes a stationary car which, 5 s
later, begins to move with a uniform acceleration
of 2 m s?2 in the same direction as the bus.
Determine a. the time taken for the car to
acquire the same velocity as the bus, b.
the distance travelled by the car when it is
level with the bus. Solution a. Given Use
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b. From the diagram,
Therefore
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Example 2.6
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
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Solution b. Instantaneous velocity at t 3
s, Use Thus
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Solution c. Instantaneous acceleration at t 3
s, Use Hence
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Exercise 2.2

• A speedboat moving at 30.0 m s-1 approaches
  stationary buoy marker 100 m ahead. The pilot
  slows the boat with a constant acceleration of
  -3.50 m s-2 by reducing the throttle.
• a. How long does it take the boat to reach the
  buoy?
• b. What is the velocity of the boat when it
  reaches the buoy?
• No. 23,pg. 51,Physics for scientists and
  engineers with modern physics, Serway
  Jewett,6th edition.
• ANS. 4.53 s 14.1 m s?1
• 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 cars 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

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Exercise 2.2

• 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
• 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

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Learning Outcome
2.3 Freely falling bodies (1 hour)

• At the end of this chapter, students should be
  able to
• Describe and use equations for freely falling
  bodies.
• For upward and downward motion, use
• a ?g ?9.81 m s?2

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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 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 s?2
• the direction is towards the centre of the earth
  (downward).
• Note
• In solving any problem involves freely falling
  bodies or free fall motion, the assumption made
  is ignore the air resistance.

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

From the sign convention thus,
Linear motion Freely falling bodies



Table 2.1
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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 decreases to zero
  when the ball reaches the maximum height, H.

velocity 0
u
Figure 2.5
v
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• 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,
  

Simulation 2.4
Figure 2.6
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• To calculate the maximum height or displacement,
  H
• use either
• maximum height,
• Another form of freely falling bodies expressions
  are

Where s H
OR
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Example 2.7
A ball is thrown from the top of a building is
given an initial velocity of 10.0 m s?1 straight
upward. The building is 30.0 m high and the ball
just misses the edge of the roof on its way down,
as shown in figure 2.7. Calculate a. the maximum
height of the stone from point A. b. the time
taken from point A to C. c. the time taken from
point A to D. d. the velocity of the ball when it
reaches point D. (Given g 9.81 m s?2)
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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
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Solution c. From point A to D, the vertical
displacement, sy ?30.0 m thus By using
c
a
b
Time dont have negative value.
OR
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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
OR
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Example 2.8
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
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Solution b. The books velocity is given
by Therefore the books
velocity is
OR
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Exercise 2.3

• 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 balls speed when it reaches the ground.
• ANS. 1.79 s 25.6 m s?1
• 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

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Exercise 2.3

• 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 balls speed when it reaches the ground.
• ANS. 1.79 s 25.6 m s?1
• 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

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Learning Outcomes
2.4 Projectile motion (2 hours)

• At the end of this chapter, students should be
  able to
• Describe and use equations for projectile,
• Calculate time of flight, maximum height, range
  and maximum range, instantaneous position and
  velocity.

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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.

Simulation 2.5
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• 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

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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
Table 2.2
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2.4.1 Maximum height, H

• 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

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2.4.2 Time taken to reach maximum height, ?t

• At maximum height, H
• Time, t ?t and vy 0
• Use

2.4.3 Flight time, ?t (from point A to point C)
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2.4.4 Horizontal range, R and value of R maximum

• 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

and
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• From the trigonometry identity,
• thus
• The value of R maximum when ? 45? and sin 2?
  1 therefore

Simulation 2.6
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2.4.5 Horizontal projectile

• 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.

Figure 2.10
Simulation 2.7
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• 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

(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
  negligible.

and
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Example 2.9
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.
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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
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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)
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Solution a. ii. magnitude and direction of
balls velocity at t 2.0 s , Horizontal
component Vertical component
Magnitude, Direction,
from positive x-axis anticlockwise
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Solution b. i. At the maximum height, H
Thus the time taken to reach maximum height
is given by ii. Apply
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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
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Solution Magnitude, Direction, therefor
e 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
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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
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Solution Magnitude, Direction, ther
efore the direction of balls velocity is
from positive x-axis anticlockwise
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Example 2.10
A transport plane travelling at a constant
velocity of 50 m s?1 at an altitude of 300 m
releases a parcel when directly above a point X
on level ground. Calculate a. the flight time of
the parcel, b. the velocity of impact of the
parcel, c. the distance from X to the point of
impact. (Given g 9.81 m s-2) Solution
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Solution The parcels velocity planes
velocity thus a. The vertical displacement is
given by Thus the flight time of the parcel is
and
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Solution b. The components of velocity of
impact of the parcel Horizontal component
Vertical component Magnitude, Directi
on, therefore the direction of parcels
velocity is
from positive x-axis anticlockwise
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Solution c. Let the distance from X to the
point of impact is d. Thus the distance, d is
given by
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Exercise 2.4

• Use gravitational acceleration, g 9.81 m s?2
• A basketball player who is 2.00 m tall is
  standing on the floor 10.0 m from the basket, as
  in Figure 2.13. If he shoots the ball at a 40.0?
  angle above the horizontal, at what initial speed
  must he throw so that it goes through the hoop
  without striking the backboard? The basket
  height is 3.05 m.
• ANS. 10.7 m s?1

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Exercise 2.4

• An apple is thrown at an angle of 30? above the
  horizontal from the top of a building 20 m high.
  Its initial speed is 40 m s?1. Calculate
• a. the time taken for the apple to strikes the
  ground,
• b. the distance from the foot of the building
  will it strikes the ground,
• c. the maximum height reached by the apple from
  the ground.
• ANS. 4.90 s 170 m 40.4 m
• 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.

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