Assignment 16
……………………………..
Projectile Motion
Name
For all questions, assume air resistance is negligible. Use “g” = 10ms-2
1. Amelia throws a netball vertically. What happens to the force on the
ball as it rises, stops then falls?
Force due to gravity is constant (downwards) the whole time
2. Amelia throws a netball vertically. What happens to the acceleration of
the ball as it rises, stops then falls?
acceleration due to gravity force is constant (downwards) the whole time (yes
even at the highest point)
3. Amelia throws a netball vertically. What happens to the velocity of the
ball as it rises, stops then falls?
velocity starts maximum upwards, decreases to zero (at highest point) then
increases but downwards.
4. What is the size and direction of the acceleration as it rises and at the
highest point?
9.8ms-2 downwards
9.8ms-2 downwards
5. Describe what 10 ms-2 actually means.
An increase or decrease of velocity by 10ms-1 in one second.
6. Penny is rock climbing. She drops a carabineer. How fast is it falling
after (a) 1 s? (b) 5 s?
(a)
(b)
10 ms-1
50 ms-1
7. Bridget throws a netball horizontally at 10 ms-1 and drops another from
the same height. Explain which lands first.
Both land at the same time because the both have the same vertical
acceleration. (sideways motion doesn’t affect vertical motion)
8. Bridget throws a netball horizontally at 10 ms-1 from a
height of 1.8 m.
(a) How long does it take to land?
(b)
How far does it go horizontally
(a) consider vertical motion.
d  vi t  1 2 at 2
1.8  0  1 2  9.8  t 2
t  0.60s
(b)consider horizontal motion
dH = v x t d = 10 x 0.60
d = 6.0 m.
9. Bridget throws the netball horizontally again from a height of 1.8 m. It
lands 4.0 m away. How fast did she throw it?
d  vi t  1 2 at 2
1.8  0  1 2  9.8  t 2
t  0.60s
dH
t
4.0
vH 
 6.7 ms 1
0.60
vH 
10.
What happens to the horizontal component of the ball’s velocity as
it falls?
Horizontal component of velocity always remains constant as there is no
horizontal force!
11. What happens to the vertical component of the ball’s velocity as it
falls?
Vertical component of velocity increases at 10 ms-2due to gravity force.
12.
Bridget then throws the netball upward at an angle. What happens
to the vertical component of its velocity as it rises? As it falls?
The vertical component of velocity decreases as the stone rises, is zero as
it changes direction; the vertical velocity increases at 10 ms-1 each second
as it falls.
13.
What happens to the horizontal component of its velocity as it
rises? As it falls?
remains constant
13.
A projectile falls beneath the straight-line path it would follow if
there were no gravity. How many meters does it fall below the line if it
has been travelling for 1s? For 2s?
d = ½at2 where a = 10ms-2. After 1 s the distance is 5m. After 2 s, distance
is 20m.
14.
Does your answer to the last question depend on the angle at
which it was launched?
NO! Acceleration remains the same.
15.
A projectile is launched vertically at 100 ms-1. If air resistance can
be neglected, at what velocity will it return to its initial level?
-100 ms-1.
16.
Suppose you roll a ball off a tabletop. Will the time to hit the
floor depend on the speed of the ball? Explain your answer.
No. Vertical motion is independent of horizontal motion. The distance
covered by the ball is given by d = ½at2, so the time taken is independent
of the initial horizontal velocity.
17.
If you toss a ball vertically upward in a uniformly moving train,
where will it land? Explain why.
It will land in the same place on the train. The ball has the same sideways
velocity as the train. If it is thrown vertically, it will keep the same
sideways velocity as the train as it has no sideways force on it.
18.
A bomb accidentally falls from a high-flying airplane just as it
flies directly above a shiny red Porsche parked in a carpark. Relative to
the Porsche, where will the bomb land?
The bomb will land past the Porsche. Distance depends on height and speed
of plane.
19.
In the absence of air drag, why does the vertical component of
velocity for a projectile change with time, whereas the horizontal
component of velocity doesn’t?
There is a vertical force (i.e. gravity) but no horizontal force, therefore
there will be no change in the horizontal velocity (Newton’s 1st Law)
20.
A friend claims that bullets fired by some high-powered rifles
travel for many meters in a straight line path before they start to fall.
Another friend disputes this claim and states that all bullets from any
rifle drop below the straight line path a vertical distance given by
d = ½at2 and that the curved path is obvious for low velocities and less
obvious for high velocities. Now it’s your turn. Will all bullets drop the
same vertical distance in equal times? Explain.
In the absence of other forces, any object falls at a rate of 10 ms-2 due to
gravity, so all objects would fall the same distance below the straight line
path in a given time. However, a slow object will cover a shorter horizontal
distance in the same time, so it’s trajectory will be more curved than that of
a faster object.
21.
An aeroplane is flying horizontally with a speed of 1000 kmh-1 (280
ms-1) when an engine falls off. Neglecting air resistance, assume it takes
30 s for the engine to hit the ground.
(a) How high is the plane?
d = ½at2 = 0.5 x 10 x 302 = 4 500 m
(b) How far horizontally does the engine travel while it falls?
d = v x t = 280 x 30 = 8 400 m
22.
If the aeroplane somehow continued to fly as if nothing had
happened, where is the engine relative to the aeroplane at the moment
the engine hits the ground?
Directly below (has the same horizontal velocity as the aeroplane).
23.
Morgan and Anna look from their 80 m high-rise balcony to a
swimming pool below, 20 m out from the bottom of their building. They
wonder how fast they would have to jump horizontally to succeed in
reaching the pool. What is the answer?
Vertical motion: t = √(2d/g) = (2 x 80/10) = 4 s
Horizontal motion: d = v x t => v = d/t = 20/4 = 5 ms-1
24.
What is the maximum speed possible for a horizontally moving
tennis ball as it clears the net 1.0 m high and strikes within the court’s
border, 12.0 m distant?
Vertical motion: t = √(2d/g) = (2 x 1/10) = √0.2 s
Horizontal motion: v = d/t = 12/0.45 = 27 ms-1
.
25.
Draw labelled velocity and acceleration vectors on the ball in
flight.
v
v
a
a
v
a
26.
Sarah kicks a soccer ball at a speed of 20 ms-1 at 30o to the
ground.
20ms-1
a. Calculate the horizontal component of velocity.
adj
hyp
v
cos 30o  x
20
vx  17ms 1
cos  
30o
Vx
b. Calculate the initial vertical component of velocity.
opp
hyp
vy
o
sin 30 
20
v y  10ms 1
sin  
c. Calculate the flight time.
time taken is the same as if it was thrown straight up at 10ms-1
find the time to get to highest point. Multiply by 2
Vy
v f  vi  at
0  10  (9.8  t )
t  1.0 s
total time  2.0s
d. Calculate how high it goes
to find height, use vertical component of velocity
2
1
2
i
d vt 
at
d  10 1  1 2  9.8 12
d  5.1m
e. Calculate how far it goes
to find distance use horizontal component of velocity
d  vH t  17  2  35m
27.
Shalini kicks a soccer ball across the length of the school pool at a
speed of 15 ms-1 at 45o to the ground. Does it reach the other end or
does it fall in the water? Justify your answer.
vx
15
vx  10.6ms 1
cos 45o 
sin 45o 
vy
15
v y  10.6ms 1
time to highest point....
use  v f  vi  at
15ms-1
Vy
45o
Vx
0  10.6  9.8  t
t  1.06ms 1
total time  2.12s
use horizontal velocity componenet to find distance
d H  vH  t
d H  10.6  2.12
d H  22m.
NO it falls in the water (assuming pool is 25m)
28.
Rose is on a tramping trip. Her friend is across a river and she
wants to throw a cell phone to her. The river is 30 m wide.
a. At what angle should she throw it?
o
45 for maximum range
b. Show that the minimum speed she can throw it is 17 ms-1
consider the motion to the highest point.
The river is 30 m so at the highest point it must have moved 15m horizontally
horizontal
vertical
d h  vh  t
vvf  vvi  at
15  vh  t
0  vv  9.8  t
t
vv
9.8
combine above equations
v
Vy
45
o
Vx
vv
)
9.8
for an angle of 450 v h  vv
15  vh (
vh2  15  9.8
v 2  vv2  vh2
v  300
v  17 ms 1
SUMMARY
Projectile motion
 no horizontal force so horizontal velocity is constant
 vertical component is the same as for an object falling down vertically –
the faster the object falls the greater the distance covered each second
 curved path followed by a projectile is a combination of horizontal and
vertical motion
Projectiles launched horizontally
 horizontal velocity is constant so ball travels same horizontal distance each
second
 ball accelerates downwards due to gravity so ball covers larger vertical
distance each second
 trajectory is a parabola
Projectiles launched at an angle
 projectile drops a distance of 5t2 below the straight line path each second
 horizontal component of velocity is constant so ball covers equal horizontal
distance each second
 vertical component of velocity changes – decreases on way up; zero at
zenith; increases on way down
 complementary angles give same horizontal range
 smaller angle => less time in the air => less range
 maximum range for angle of 45º