PROJECTILE MOTION PRACTICE QUESTIONS (WITH ANSWERS) * challenge questions
2 views 0 purchase
Course
PROJECTILE MOTION
Institution
PROJECTILE MOTION
PROJECTILE MOTION
PRACTICE QUESTIONS (WITH ANSWERS)
* challenge questions
Q1.
A golfer practising on a range with an elevated tee
4.9 m above the fairway is able to strike a ball so
that it leaves the club with a horizontal velocity of
20 m s–1. (Assume the acceleration due to gra...
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?
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller StudyCenter1. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $13.49. You're not tied to anything after your purchase.
