PROJECTILE MOTION e The ball will strike the ground 1.0 s after it is
PRACTICE QUESTIONS (WITH ANSWERS) struck.
Then vx = 20 m s–1
* challenge questions
and vy = 0 + (9.8 m s–2)(1.0 s) = 9.8 m s–1
The speed of the ball at 1.0 s is given by:
[(20 m s–1)2 + (9.8 m s–1)2] ½ = 22.3 m s–1
Q2.
Q1.
A bowling ball of mass 7.5 kg travelling at 10 m s–1
A golfer practising on a range with an elevated tee
rolls off a horizontal table 1.0 m high. (Assume the
4.9 m above the fairway is able to strike a ball so
acceleration due to gravity is 9.80 m s–2, and the
that it leaves the club with a horizontal velocity of
effects of air resistance may be ignored unless
20 m s–1. (Assume the acceleration due to gravity is
otherwise stated.)
9.80 m s–2, and the effects of air resistance may be
a Calculate the ball’s horizontal velocity just as it
ignored unless otherwise stated.)
strikes the floor.
b What is the vertical velocity of the ball as it
strikes the floor?
c Calculate the velocity of the ball as it reaches
the floor.
d What time interval has elapsed between the
ball leaving the table and striking the floor?
e Calculate the horizontal distance travelled by
the ball as it falls.
a How long after the ball leaves the club will it
land on the fairway? A2.
b What horizontal distance will the ball travel a The horizontal velocity of the ball remains
before striking the fairway? constant and vx = 10 m s–1.
c What is the acceleration of the ball 0.5 s after b v2 = u2 + 2ax
being hit? and vy2 = 02 + 2(9.8 m s–2)(1.0 m)
d Calculate the speed of the ball 0.80 s after it and vy = 4.4 m s–1 down
leaves the club. c v = [(10 m s–1)2 + (4.43 m s–1)2] ½ = 10.9 m s–1 at
e With what speed will the ball strike the 24° to the horizontal,
ground? where the angle is determined from
tan θ = 4.43 m s–1/10 m s–1 = 0.443 and θ = 24°
A1. d x = ut + 0.5at2
a x = ut + 0.5at2 and 1.0 m = 0 + 0.5(9.8 m s–2)t2
then 4.9 m = 0 + 0.5(9.8 m s–2)t2 so t = 0.45 s
and t = 1.0 s e Horizontal distance = (horizontal speed)(time)
b x = (average speed)(time) = (20 m s–1)(1.0 s) = = (10 m s–1)(0.45 s) = 4.5 m
20 m
c The acceleration of the ball is constant at any
time during its flight, and is equal to the
acceleration due to gravity
= 9.8 m s–2 down
d After 0.80 s, the ball has two components of
velocity:
vx = 20 m s–1
and vy = 0 + (9.8 m s–2)(0.80 s) = 7.84 m s–1
The speed of the ball at 0.80 s is given by:
[(20 m s–1)2 + (7.84 m s–1)2]½ = 21.5 m s–1
, QUESTIONS 3 - 8
A senior physics class conducting a research A5.
project on projectile motion constructs a device a The time for the ball to reach its maximum
that can launch a cricket ball. The launching device height is determined from v = u + at.
is designed so that the ball can be launched at Then at maximum height, the vertical velocity
ground level with an initial velocity of 28 m s–1 at of the ball = 0
an angle of 30° to the horizontal. and 0 = 14 m s–1 – (9.8 m s–2)t
and t = 1.43 s
b v2 = u2 + 2ax
then 0 = (14 m s–1)2 – (9.8 m s–2)x
and x = 10 m
c The acceleration of the ball is constant at any
Q3.
time during its flight, and is equal to the
Calculate the horizontal component of the velocity
acceleration due to gravity = 9.8 m s–2 down.
of the ball:
a initially
Q6.
b after 1.0 s
a At which point in its flight will the ball
c after 2.0 s.
experience its minimum speed?
b What is the minimum speed of the ball during
A3.
its flight?
a vx = (28 m s–1) cos 30° = 24.2 m s–1 north and
c At what time does this minimum speed occur?
remains constant throughout the flight.
b 24.2 m s–1 north
A6.
c 24.2 m s–1 north
a The minimum speed will occur when the
vertical components of the ball’s velocity = 0,
Q4.
i.e. at the maximum height.
Calculate the vertical component of the velocity of
b The minimum velocity of the ball during its
the ball:
flight occurs at the maximum height, and is
a initially
equal to the horizontal component of the
b after 1.0 s
ball’s velocity = 24.2 m s–1 horizontally.
c after 2.0 s.
c The minimum speed of the ball during its flight
occurs at the maximum height at t = 1.43 s.
A4.
a vy = (28 m s–1) sin 30° = 14 m s–1 up
Q7.
b vy = 14 m s–1 – (9.8 m s–2)(1.0 s) = 4.2 m s–1 up
a At what time after being launched will the ball
c The time for the ball to reach its maximum
return to the ground?
height is determined from v = u + at.
b What is the velocity of the ball as it strikes the
Then at maximum height, the vertical velocity
ground?
of the ball = 0
c Calculate the horizontal range of the ball.
and 0 = 14 m s–1 – (9.8 m s–2)t
and t = 1.43 s
A7.
Therefore at t = 2.0 s the ball is 0.57 s into its
a The flight of the ball is symmetrical. Therefore
downward flight.
the time for it to reach the ground after
vy = 0 + (9.8 m s–2)(0.57 s) = 5.6 m s–1 down
launching = 2(1.43 s) = 2.86 s.
b The flight of the ball is symmetrical. Therefore
Q5.
the ball will strike the ground at the same
a At what time will the ball reach its maximum
velocity as that when it was launched: 28 m s–1
height?
at an angle of 30° to the horizontal.
b What is the maximum height that is achieved
c Horizontal range = (horizontal speed)(time)
by the ball?
= (24.2 m s–1)(2.86 s) = 69.2 m
c What is the acceleration of the ball at its
maximum height?
PRACTICE QUESTIONS (WITH ANSWERS) struck.
Then vx = 20 m s–1
* challenge questions
and vy = 0 + (9.8 m s–2)(1.0 s) = 9.8 m s–1
The speed of the ball at 1.0 s is given by:
[(20 m s–1)2 + (9.8 m s–1)2] ½ = 22.3 m s–1
Q2.
Q1.
A bowling ball of mass 7.5 kg travelling at 10 m s–1
A golfer practising on a range with an elevated tee
rolls off a horizontal table 1.0 m high. (Assume the
4.9 m above the fairway is able to strike a ball so
acceleration due to gravity is 9.80 m s–2, and the
that it leaves the club with a horizontal velocity of
effects of air resistance may be ignored unless
20 m s–1. (Assume the acceleration due to gravity is
otherwise stated.)
9.80 m s–2, and the effects of air resistance may be
a Calculate the ball’s horizontal velocity just as it
ignored unless otherwise stated.)
strikes the floor.
b What is the vertical velocity of the ball as it
strikes the floor?
c Calculate the velocity of the ball as it reaches
the floor.
d What time interval has elapsed between the
ball leaving the table and striking the floor?
e Calculate the horizontal distance travelled by
the ball as it falls.
a How long after the ball leaves the club will it
land on the fairway? A2.
b What horizontal distance will the ball travel a The horizontal velocity of the ball remains
before striking the fairway? constant and vx = 10 m s–1.
c What is the acceleration of the ball 0.5 s after b v2 = u2 + 2ax
being hit? and vy2 = 02 + 2(9.8 m s–2)(1.0 m)
d Calculate the speed of the ball 0.80 s after it and vy = 4.4 m s–1 down
leaves the club. c v = [(10 m s–1)2 + (4.43 m s–1)2] ½ = 10.9 m s–1 at
e With what speed will the ball strike the 24° to the horizontal,
ground? where the angle is determined from
tan θ = 4.43 m s–1/10 m s–1 = 0.443 and θ = 24°
A1. d x = ut + 0.5at2
a x = ut + 0.5at2 and 1.0 m = 0 + 0.5(9.8 m s–2)t2
then 4.9 m = 0 + 0.5(9.8 m s–2)t2 so t = 0.45 s
and t = 1.0 s e Horizontal distance = (horizontal speed)(time)
b x = (average speed)(time) = (20 m s–1)(1.0 s) = = (10 m s–1)(0.45 s) = 4.5 m
20 m
c The acceleration of the ball is constant at any
time during its flight, and is equal to the
acceleration due to gravity
= 9.8 m s–2 down
d After 0.80 s, the ball has two components of
velocity:
vx = 20 m s–1
and vy = 0 + (9.8 m s–2)(0.80 s) = 7.84 m s–1
The speed of the ball at 0.80 s is given by:
[(20 m s–1)2 + (7.84 m s–1)2]½ = 21.5 m s–1
, QUESTIONS 3 - 8
A senior physics class conducting a research A5.
project on projectile motion constructs a device a The time for the ball to reach its maximum
that can launch a cricket ball. The launching device height is determined from v = u + at.
is designed so that the ball can be launched at Then at maximum height, the vertical velocity
ground level with an initial velocity of 28 m s–1 at of the ball = 0
an angle of 30° to the horizontal. and 0 = 14 m s–1 – (9.8 m s–2)t
and t = 1.43 s
b v2 = u2 + 2ax
then 0 = (14 m s–1)2 – (9.8 m s–2)x
and x = 10 m
c The acceleration of the ball is constant at any
Q3.
time during its flight, and is equal to the
Calculate the horizontal component of the velocity
acceleration due to gravity = 9.8 m s–2 down.
of the ball:
a initially
Q6.
b after 1.0 s
a At which point in its flight will the ball
c after 2.0 s.
experience its minimum speed?
b What is the minimum speed of the ball during
A3.
its flight?
a vx = (28 m s–1) cos 30° = 24.2 m s–1 north and
c At what time does this minimum speed occur?
remains constant throughout the flight.
b 24.2 m s–1 north
A6.
c 24.2 m s–1 north
a The minimum speed will occur when the
vertical components of the ball’s velocity = 0,
Q4.
i.e. at the maximum height.
Calculate the vertical component of the velocity of
b The minimum velocity of the ball during its
the ball:
flight occurs at the maximum height, and is
a initially
equal to the horizontal component of the
b after 1.0 s
ball’s velocity = 24.2 m s–1 horizontally.
c after 2.0 s.
c The minimum speed of the ball during its flight
occurs at the maximum height at t = 1.43 s.
A4.
a vy = (28 m s–1) sin 30° = 14 m s–1 up
Q7.
b vy = 14 m s–1 – (9.8 m s–2)(1.0 s) = 4.2 m s–1 up
a At what time after being launched will the ball
c The time for the ball to reach its maximum
return to the ground?
height is determined from v = u + at.
b What is the velocity of the ball as it strikes the
Then at maximum height, the vertical velocity
ground?
of the ball = 0
c Calculate the horizontal range of the ball.
and 0 = 14 m s–1 – (9.8 m s–2)t
and t = 1.43 s
A7.
Therefore at t = 2.0 s the ball is 0.57 s into its
a The flight of the ball is symmetrical. Therefore
downward flight.
the time for it to reach the ground after
vy = 0 + (9.8 m s–2)(0.57 s) = 5.6 m s–1 down
launching = 2(1.43 s) = 2.86 s.
b The flight of the ball is symmetrical. Therefore
Q5.
the ball will strike the ground at the same
a At what time will the ball reach its maximum
velocity as that when it was launched: 28 m s–1
height?
at an angle of 30° to the horizontal.
b What is the maximum height that is achieved
c Horizontal range = (horizontal speed)(time)
by the ball?
= (24.2 m s–1)(2.86 s) = 69.2 m
c What is the acceleration of the ball at its
maximum height?
