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Knowledge and understanding
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When you have finished this chapter, you should be able to:
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•
•
describe the trajectory of an object undergoing projectile motion
within the Earth’s gravitational field in terms of horizontal and
vertical components
solve problems and analyse information to calculate the actual
velocity of a projectile from its horizontal and vertical components
using the equations:
•
2
Projectiles
v2x u2x
v u at
v2y u2y 2ay y
x ux t
1 a t2
y uy t __
2 y
describe Galileo’s analysis of projectile motion.
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Figure 2.1 Projectile motion is exhibited by fireworks. (The effects of air resistance alter the paths of
some projectiles.)
Figure 2.2 A skier undergoing projectile motion.
2.1 Projectile motion
projectile motion
The movement of objects through
space or the atmosphere, moving
freely under the influence of
gravity
Projectile motion refers to the movement of objects through space or the
atmosphere, moving freely under the influence of gravity. It includes a
coin being dropped, a ball being kicked, a bullet being fired and satellites
in orbit around a planet. To study the motion of projectiles, it is necessary
to break the motion up into its vector components and consider the
horizontal motion separately to the vertical motion.
Consider an object moving under the force of gravity. If it is a coin
dropped from a height of, say, 1 metre, then it will start its motion with a
velocity of zero and will gain in speed, as it accelerates in the Earth’s
gravitational field, until it hits the ground and comes to rest. If, on the
other hand, it is flipped upwards, it will start its motion with the velocity
imparted to it in the flip, and will gradually slow down under the influence
of gravity until it reaches a velocity of zero, when it will start to accelerate
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Module 1
Chapter 2 \\ Projectiles
downwards, gaining speed until it reaches the ground and comes to rest.
The only force acting on the motion of the coins in the two situations is
the constant force of gravity. If the coin is now flicked off a tabletop that is
1 metre above the ground, so that horizontal motion is imparted to it, will
it reach the ground at the same time as the dropped coin? What if a coin
and an apple are dropped from the same height – will they reach the
ground at the same time? Galileo found that they did. Legend has him
dropping cannonballs of different masses off the Tower of Pisa while
employed by the University of Pisa (around 1590) to confirm that they did
indeed reach the ground at the same time. This was the start of our modern
study of kinematics – the motion of objects along a trajectory.
In Aristotle’s theory of projectile motion, projectiles were pushed along
by an external force called ‘impetus’ that was transmitted through the air.
This impetus caused the object to move in a straight line until the force
was expended, at which point the object fell to the ground. With the spread
of cannon warfare, the study of projectile motion took on a greater
importance and it was realised through careful observation that projectiles
do not behave in this Aristotelian manner. It was realised that projectiles
followed the path of some sort of smooth curve, but it needed further
study to determine the shape of this curve.
Galileo had been studying balls rolling down smooth inclined planes to
show that the force acting on them (gravity) was a constant and that no
matter the mass of the balls, they rolled down the plane with the same
acceleration. In other words, the time taken for balls of different masses to
travel down the inclined plane was independent of their mass and only
dependent on the force acting on them.
Galileo now took his studies one step further. He allowed pre-inked
balls to roll down an inclined plane on a table and off the edge of the table.
The balls left a mark on a sheet of paper on the floor, indicating their
landing position. This mark allowed the horizontal and vertical distances
travelled by the ball to be measured. By varying the horizontal velocity
and the vertical drop of the ball, Galileo was able to determine that the
path of a projectile is parabolic. Galileo had reasoned that a projectile
shot from a cannon is influenced not by one motion only, but by two at
right angles to each other. Galileo was able to demonstrate using his
inclined planes that a projectile is influenced by two independent
motions – one horizontal and the other vertical and influenced by gravity.
13
Projectile motion
kinematics
The study of the motion of objects
along a trajectory without regard
to its causes
Figure 2.3 The two independent motions – vertical
and vertical combined with the horizontal.
2.2 Equations of motion
There are three equations of motion that we can apply to our study of
projectiles. They describe the motion of objects under the influence of a
gravitational field, no matter how strong or weak that field is. These
equations also assume that there is no wind or air resistance, which is
reasonable in most instances.
v u at
v 2 u 2 2ar
1 at 2
r ut __
2
where
v final velocity (m s1)
u initial velocity (m s1)
t time (s)
a acceleration (m s2)
r displacement (m)
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The equations can be applied to any situation where the acceleration
is constant or zero. They can be used to quantitatively describe the
motion of a car accelerating at 0.5 m s1, or a tennis ball as it follows its
trajectory through the air under the influence of gravity. In both these
situations, the acceleration that is applied may be considered to be
constant. Gravity on Earth is taken to be a constant 9.8 m s2 downwards
as the differences in g at the Earth’s surface to do with latitude and
altitude are minimal.
Let’s go back to our coin and this time consider the situation in which it
is dropped from a height of 5 m.
\\ WORKED EXAMPLE
Question 1
A coin is dropped from a height of 5.0 m.
a At what speed is it travelling when it hits the ground?
b How long does it take before it hits the ground?
Answer
Now it becomes very important to identify all of the data in the question.
Data: a g 9.8 m s2; r 5.0 m; u 0 m s1; t ?; v ? (Note that
a or – simply defines the direction of the quantity.)
Knowing a, r and u leads us to the equation v 2 u 2 2ar to answer part a.
v 2 u 2 2ar
0
2 9.8 5
____
v 98.0
9.9 m s1
a
b
Velocity at the point of impact is 9.9 m s1 downwards. (Our knowledge of
gravity tells us that it must be downwards.)
Data: a g 9.8 m s2; r 5.0 m; u 0 m s1; v 9.9 m s1;
t?
v u at
9.9 0 9.8t
9.9
t ___
9.8
1.0 s
The time taken for the coin to drop is 1.0 s.
Question 2
An object is thrown vertically in the air at a velocity of 50 m s1. Calculate:
a how high the object goes
b the object’s velocity at 6 seconds
c the object’s position at 6 seconds.
Answer
Start with the data.
a Data: a g 9.8 m s2; u 50 m s1; at its highest point it will have
a velocity of zero, so v 0; r ?
v 2 u 2 2ar
v 2 u2
r _______
2a
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Chapter 2 \\ Projectiles
0 2500
r ________
19.6
127.6 m
b
c
15
Equations of motion
Highest point reached is 127.6 m above its starting point.
Data: a g 9.8 m s2; u 50 m s1; t 6 s; v ?
v u at
50 9.8 6
8.8 m s1
At time 6 seconds the object is travelling with a velocity of 8.8 m
downwards.
Data: a g 9.8 m s2; u 50 m s1; v 8.8 m s1; t 6 s; r ?
1at 2
r ut __
2
1(9.8 6 2)
(50 6) __
2
300 176.4
123.6 m
The object will be at a position 123.6 m above its starting point.
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Equations of motion
The first equation of motion is simply a rearrangement of the equation for
acceleration studied in Preliminary Physics:
change in velocity
average acceleration ________________
time taken
v
u
______
a
t
at v u
v u at
(1)
A second equation begins with the expression for average velocity studied
in Preliminary Physics:
total displacement
average velocity _________________
time
r
_
vav t
Average velocity can also be found by summing the individual velocities:
uv
vav _____
2
So now we have:
u v _r
_____
t
2
Substituting in v u at, gives:
u (u at ) _r
___________
t
2
2u at _r
_______
2
t
1 at 2
r ut __
2
The last equation is derived from the first two equations:
v u at
1 at 2
r ut __
2
(2)
(1)
(2)
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From equation (1):
vu
t ______
a
(1a)
Substituting equation (1a) into (2) gives:
v u __
vu 2
1 ______
r u ______
a 2 a a a(v u)(v u)
uv u 2 __
1 ______________
r ________
a
2
a2
1v 2 __
1u 2 uv
ar uv u 2 __
2
2
1
1
__
__
2
2
ar v u
2
2
v 2 u 2 2ar
(3)
This is the third equation of motion.
2.3 Motion in two dimensions
Examining projectile motion,
pages 12–16, Practical Physics
for Senior Students, HSC
g
m
u
Figure 2.4 The motion of an object of mass m
moving under gravity with an initial horizontal
velocity u៝.
Consider a marble rolling off a table with an initial horizontal velocity
(Figure 2.4). Its initial vertical velocity is 0, but it is subject to gravitational
acceleration of g downwards (g). (By convention, upwards is taken as
positive.) The motion of this object can be described by taking components
parallel to and at right angles to the direction of the constant force acting on
it (in this case g). In fact, the trajectory taken by the marble is that of a
parabola; however, we break it up and consider its horizontal and vertical
components separately.
Gravity only acts on an object in the vertical direction. The horizontal
motion of the marble will behave exactly as if it were allowed to roll along
a smooth and frictionless surface. Its initial velocity will be the same as its
final velocity. There is no acceleration acting on it in the horizontal
direction. Its motion simply comes to a halt, because that is determined
by what is happening in the vertical component of its motion.
In the horizontal direction we find that:
ux ux (initial horizontal velocity)
ax 0 (as no force acts in this direction)
rx x (horizontal displacement)
The three equations of motion become:
vx ux
(1)
vx2 ux2
(2)
x u xt
(3)
In the vertical direction we find that:
uy 0 (initially not moving in the vertical direction)
ay g (acceleration due to gravity)
ry y (vertical displacement)
The three equations of motion become:
vy gt
(1)
(2)
vy 2gr
1
(3)
y __gt 2
2
If the marble were projected horizontally with a speed of 3.0 m s1, the
horizontal and vertical components are as shown on the graphs in Figure 2.5.
2
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17
1
4
vh = 3.0 m s–1
2
1
2
3
Time (s)
Displacement (m)
a=0
Velocity (m s–1)
Acceleration
(a) Horizontal components
9
6
x = 3.0
3
2
3
Time (s)
1
2
3
Time (s)
a = 9.8 m s–2
10
5
1
2
3
Time (s)
20
vv = 9.8 t
10
1
2
3
Time (s)
Displacement (m)
30
Velocity (m s–1)
Acceleration (m s–2)
(b) Vertical components
45
30
y = 4.9 t 2
15
1
2
3
Time (s)
Figure 2.5 (a) The horizontal and (b) the vertical components for the path of a ball projected horizontally with a
speed of 3.0 m s1.
\\ WORKED EXAMPLE
Question 3
An object is projected horizontally with a velocity of 5.0 m s1.
a When will its vertical displacement be 20 m below its starting point?
b What will its horizontal displacement be at this time?
c Find its velocity at this time.
Answer
a
b
Vertically:
Data: a g 9.8 m s2; u 0 m s1; y 20 m; t ? (Remember
that downwards is negative.)
This data enables the use of r ut _12 at 2 to find t.
Two-dimensional projectile
motion
1(9.8)t 2
20 0 __
2
20
t 2 _____
4.9
___
t 4.1
t 2.0 s
The object will be 20 m below its starting point at 2.0 s.
Horizontally:
Data: a 0 m s2; ux vx 5.0 m s1; t 2.0 s; x ?
This data enables the use of r ut _12 at 2 to find x, and in the horizontal
simplifies to:
x ut
5.0 2.0
10.0 m
The object has travelled a distance of 10.0 m in the horizontal.
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c
5 m s−1
v
19.6 m s−1
The velocity is the vector sum of the horizontal and vertical components.
Horizontally:
vx 5.0 m s1
Vertically:
vy uy ayt
vy 0 9.8 2.0
19.6 m s1
From the vector diagram (Figure 2.6):
_______
Figure 2.6 Vector addition triangle.
v vx 2 vy 2
___________
5.0 2 19.6 2
______
409.16
20.2 m s1
at an angle of:
19.6
tan ____
5
76
The resultant velocity of the combined components is 20.2 m s1 at an
angle of 76° below the horizontal.
Question 4
A rock is thrown horizontally from an 80 m high cliff. It lands at a point on level
ground 40 m from the base of the cliff.
a How long does the rock take to reach the ground below the cliff?
b At what speed was it thrown?
Answer
cliff
80 m
rock
40 m
Figure 2.7 A rock 40 m from base of an 80 m high cliff.
It is often helpful to draw a diagram of the problem.
a Vertically:
Data: a g 9.8 m s2; u 0 m s1; y 80 m; t ?
1at 2 to find t.
This data enables us to use r ut __
2
1gt 2
y uyt __
2
1 (9.8)t 2
80 0 __
2
80
t 2 _____
4.9
____
t 16.3
4.0 s
b
The time taken to reach the ground is 4.0 s.
Horizontally:
Data:
a 0 m s1; t 4.0 s; x 40 m; ux vy ?
1at 2, which in the horizontal
This enables us to use r ut __
2
simplifies to:
x ux t
40 ux 4.0
ux 10.0 m s1
The rock was thrown horizontally with a speed of 10.0 m s1.
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Module 1
Chapter 2 \\ Projectiles
19
Problem set 2A
Question 1
A rock is dropped from a height of 3 m.
a
b
Calculate the speed that it attains.
How long is it in the air?
Question 2
Calculate the final velocity of an object with an initial velocity of
35 m s1 and an acceleration of 6 m s2 over a period of 5 s.
Question 3
If v 10 m s1, a 4 m s2 and t 12 s, find u.
Question 4
A coin is flipped directly upwards with an initial speed of 2.0 m s1.
a
b
What is the maximum height that it reaches?
How long is it in the air, assuming that it is caught at exactly the
same height from which it was released?
Question 5
An object is launched horizontally with an initial velocity of 5.0 m s1.
What will be the equations for the:
a
b
y - coordinate of its displacement after time t ?
x - coordinate of its displacement after time t ?
Question 6
An object is projected horizontally with a velocity of 2.0 m s1.
a
b
c
d
What will the horizontal displacement be after 2.0 s?
Find the vertical displacement at this time.
When will the horizontal displacement be 0.20 m?
What will the vertical displacement be when the horizontal
displacement is 0.20 m?
Question 7
An object is projected horizontally. After t seconds it has a horizontal
displacement of 10.0 cm and a vertical displacement of 5.0 cm.
a
b
What is the value of t ?
What is the magnitude of the initial velocity?
Question 8
An object is projected horizontally with an initial velocity of 3.0 m s1.
a
b
When will the vertical displacement be 15.0 m?
What will the horizontal displacement be at this time?
Question 9
A stone is thrown horizontally out to sea from the top of a cliff, with
a velocity of 15 m s1, and reaches the water below after 3.0 s.
a
How high is the cliff?
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b
c
How far from the base of the cliff is the stone when it reaches
the water?
What is the velocity of the stone just prior to striking the water?
Question 10
A multiflash photograph is taken of an object that is projected
horizontally. How can you tell from the photograph that the horizontal
velocity component is constant and that the vertical velocity
increases uniformly?
Question 11
A ball of mass 0.50 kg rolls off a horizontal ledge 4.0 m above the
ground with a speed of 4.0 m s1.
a
b
c
d
e
f
How long does it take the ball to reach the ground below the
ledge?
How far, horizontally, from the ledge will the ball strike the
ground?
What is the horizontal component of the ball’s velocity when it
strikes the ground?
What is the vertical component of the ball’s velocity when it
strikes the ground?
What is the kinetic energy of the ball just before it strikes the
ground?
What is the speed of the ball as it strikes the ground?
Question 12
Two balls, A and B, each of mass 1.6 kg are rolling along a
horizontal table 2.0 m above the floor. Both balls roll off the end of
the table and reach the floor below. A has a horizontal velocity of
4.0 m s1 as it leaves the table, while B has a horizontal velocity of
2.5 m s1.
a
b
What is the time taken for:
i
A to reach the floor?
ii
B to reach the floor?
What is the difference in the horizontal distances travelled by A
and B before they strike the floor?
2.4 Projectile motion at an
angle
Calculations of projectile
motion, pages 9–12, Practical
Physics for Senior Students,
HSC
Obviously not all projectiles are either purely vertical, such as the
dropped coin, or simple two-dimensional examples where there is
only a horizontal velocity initially imparted to the object, such as in
the case of the marble rolled off the flat table. Projectiles often start
their motion at an angle, such as when a soccer ball is kicked. The
soccer ball moves off at an angle to the horizontal and follows the
trajectory of a parabola.
Consider an object projected with speed u at an angle as shown in
Figure 2.8 (ignore air resistance).
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Chapter 2 \\ Projectiles
21
u
h
u sin θ
θ
0
u cos θ
x
θ
R
u
Figure 2.8 The motion of a projectile with an initial velocity u at an angle of to the horizontal.
Consider the vector triangle for the initial motion of this object. The
horizontal (ux ) and vertical (uy ) components of the velocity are
independent and are found as follows.
Horizontally:
ux
cos __
u
ux u cos Vertically:
uy
sin __
u
u
u sin θ
θ
u cos θ
Figure 2.9 Vector triangle for
the motion of an object with
an initial velocity u at an angle
of to the horizontal.
uy u sin So the horizontal component is given by u cos and the vertical
component by u sin , where is the angle between the initial velocity and
the horizontal.
The equations for the components of this motion are shown in Table 2.1.
Consider an object projected at 50 m s1 at such an angle that the
horizontal component of the velocity is 30 m s1 and the vertical
component is 40 m s1 in a region where the gravitational field strength is
9.8 N kg1. This means that the acceleration of a freely falling object is
9.8 m s2 downwards.
The position and velocity values can be calculated from the equations
for motion under constant acceleration.
Note the following:
a
b
c
d
Table 2.1
Horizontal
Vertical
vx ux cos vy changing
ux ux cos uy uy sin rx
ry
a0
a g
tt
tt
The horizontal velocity (component) is constant at 30 m s1.
The vertical velocity (component) changes by 9.8 m s1 each second
(down).
The minimum velocity occurs at the top it is not zero, it equals the
value of the horizontal component.
The acceleration is always 9.8 m s2 down, even at the top position,
because the force of gravity acts at all times throughout the
motion.
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\\ WORKED EXAMPLE
Question 5
A soccer ball is kicked at an angle of 37.0° to the horizontal with a velocity of
25.0 m s1 as shown in Figure 2.10. Calculate the:
a maximum height
b time of travel
c range of the soccer ball (horizontal distance travelled)
d velocity at the maximum height
e acceleration at the maximum height.
(Neglect air resistance and any spinning of the ball.)
.0
m
s–
1
25
37°
Figure 2.10
Answer
a
b
Maximum height will be due to the initial vertical velocity and occurs when
vertical velocity reaches a velocity of zero.
Vertically:
uy u sin 25.0 sin 37
15.0 m s1
vy 0
a g 9.8 m s2
y?
2
2
Using v u 2ar gives:
v 2 u2
y ______
2a
0 15.02
y ________
2(9.8)
225
______
19.6
11.5 m
The maximum height reached is 11.5 m.
The time is also found using the vertical component and is the time taken
to go up to the maximum height and back down.
Vertically:
Data: a g 9.8 m s2; uy 15.0 m s1; vy 15.0 m s1; t ?
Using v u at gives:
vu
t _____
a
15.0
15.0
____________
9.8
3.1 s
The total time of travel is 3.1 s.
Note: This could also have been found by finding the time to reach its
maximum height and doubling that result.
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Chapter 2 \\ Projectiles
c
d
e
23
To find the range (horizontal distance travelled), the horizontal component
of velocity is used in conjunction with the time taken for the object’s
journey.
Horizontally:
ux u cos 25.0 cos 37
20.0 m s1
Data: a 0 m s1; ux vx 20.0 m s1; t 3.1 s; x ?
1 at 2 gives:
Using r ut __
2
x ux t
20.0 3.1
62.0 m
The range of the soccer ball is 62.0 m.
At the maximum height, the vertical component of the velocity is zero.
There is only the constant horizontal component. Hence, the velocity at the
greatest height is 20.0 m s1 in a horizontal direction.
The acceleration at the maximum height is the same as at any other point
in the parabolic path throughout the flight, namely 9.8 m s2 downwards.
Note that even though the vertical velocity component is zero at the
maximum height, the ball is still in the Earth’s gravitational field so that it
still experiences a force of attraction.
\\ DID YOU KNOW?
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Air resistance
Air resistance is the name given to the friction force that opposes the
motion of a projectile (or anything else) through the air.
As an object moves through the air, it is constantly colliding with air
molecules, and therefore transferring energy and momentum to them. The
magnitude of this air resistance depends on the size and shape of the
projectile, and its speed.
The greater the object’s cross-sectional area at right angles to its
direction of motion, the more molecules in the air it will collide with.
Because of this, large objects experience more air resistance than do
small objects. For example, a sheet of paper falls more quickly if it is
crumpled into a ball.
More kinetic energy is lost in a head-on collision than in a glancing
collision. A streamlined object is shaped so that its leading surfaces strike
the air at an angle, deflecting the air rather than colliding head-on with it.
A streamlined object therefore experiences less air resistance than one
that is not streamlined.
The faster an object is travelling, the more air molecules it will strike in
a given time and, therefore, the greater the air resistance it will experience.
In the discussion above, we have considered an object moving
through still air, but the arguments apply equally to the force exerted on
an object by a moving air stream.
We have also considered the molecules in the air as being stationary,
but in fact they are moving around in all directions at high speed. In the
absence of wind, however, all these individual movements cancel out and
we can ignore them.
If you throw a cricket ball, a tennis ball and a table tennis ball, you will
be aware of some of the effects of air resistance. A cricket ball has a
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higher mass to surface area ratio than a tennis ball and a table tennis ball,
so it is easier to project it through the air. The path of any ball will not be
parabolic in a strong wind.
Objects dropped from moving aeroplanes would follow a parabolic
path if they were not subjected to air resistance.
The pilots of planes and helicopters dropping parcels and rescue
equipment when in flight need to make allowance for the parabolic path of
a projectile. The actual path will be affected by air resistance.
For a ball or similar body in motion through the air, there will be two
forces acting on it. The force of gravity will be acting vertically downwards
and the force of air resistance will be acting against the motion. These
forces are shown in Figure 2.11. The magnitude and the direction of the
force of air resistance will change as the direction of motion and the
speed change. However, the magnitude and direction of the gravitational
force on the body will not change.
v
Fa
mg
Fa
Fa
mg
mg
Figure 2.11 The forces acting on an object moving through the air.
Problem set 2B
Question 1
A cricket ball is thrown in the air at an angle of 60° to the horizontal
at 12 m s1.
a
b
c
d
Calculate the initial:
i
horizontal velocity
ii
vertical velocity.
Calculate the maximum height that it attains.
Find how long it is in the air before it returns back to ground
level.
Calculate its distance travelled in the horizontal.
Question 2
A projectile is launched at 55 m s1 at an angle of 30° to the
horizontal.
a
b
c
d
e
What is
Find the
What is
What is
What is
its initial horizontal velocity?
vertical component of its velocity initially.
the horizontal component of its velocity after 3 s?
the vertical component of its velocity after 3 s?
its velocity after 3 s?
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Module 1
Chapter 2 \\ Projectiles
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Question 3
A shell is fired from a cannon at 350 m s1 at an angle of 45° to the
horizontal.
a
b
c
d
e
Calculate the maximum height that it reaches.
Find the time that it is in the air.
Calculate its range.
Find the final:
i
horizontal velocity
ii
vertical velocity.
What is the velocity with which it hits the ground?
Question 4
An object is launched upwards at an angle of 55° to the horizontal
with a velocity of 80 m s1.
a
b
How long before it hits the ground again?
How high does it go?
Question 5
A projectile is launched at 140 m s1 at an angle of 50° to the
ground.
a
b
c
Where is it 5 s after launching?
Where does it strike the ground?
What is the greatest height that it reaches?
Question 6
A projectile is fired at 20 m s1 at 30° to the horizontal.
a
b
c
Calculate its time of flight.
Calculate its range.
What is the maximum height reached?
Question 7
A projectile has a range of flight of 1120 m and a time of flight of
8.0 s. Calculate the:
a
b
c
d
initial horizontal velocity
initial vertical velocity
initial velocity of the projectile
maximum height reached.
Question 8
A cannonball rises to a height of 90 m after being fired at 60° to the
horizontal. Find the:
a
b
c
initial vertical velocity
initial horizontal velocity
range of the cannonball.
Question 9
A boy standing vertically on the tray of a lorry travelling at 20 km h1
throws a ball vertically upwards with a speed of 7 m s1 and catches
it again at the same level. What distance horizontally does the ball
move while it is in the air?
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Question 10
A cricket ball is hit with a velocity of 8.0 m s1 at an angle of 60°
with the horizontal. Calculate:
a
b
its horizontal and vertical displacements after 0.50 s has
elapsed
the time taken to return to the level from which it was hit, and
the horizontal distance travelled in this time.
Question 11
How far would a ball go if it was struck at an angle of 40° from the
horizontal at 20.0 m s1?
Question 12
A shell is fired from a gun at 350 m s1 at an angle of 15° to the
horizontal. What would be the greatest height to which the shell
would rise?
Question 13
A ball is thrown a distance of 75 m at an elevation of 23°. What
was its original speed?
Question 14
On the planet Zag, an astronaut kicked a ball at an angle of 45°,
giving the ball an initial speed of 5.0 m s1. The ball flew 15 m before
hitting the ground. What is the value of g on Zag?
Question 15
A projectile is fired at 30° to the horizontal from the top of a cliff
150 m high. Its initial speed is 49 m s1.
a
Calculate its maximum height above the:
i
cliff
ii
ground.
b
c
Calculate the total time for which the projectile is in the air.
Calculate the distance from the base of the cliff that the
projectile lands.
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Chapter 2 \\ Projectiles
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Summary of projectiles
•
•
•
•
•
•
•
A projectile is any object moving under the influence of a gravitational field.
A projectile follows the path of a parabola.
Projectiles on Earth move under the force of gravity.
Horizontal motion is independent of vertical motion in projectiles.
The three equations of motion are:
v u at
v 2 u 2 2ar
r ut __1 at 2
2
Horizontal motion may be considered to have no acceleration.
Vertical motion is affected by the pull of gravity (9.8 m s2).
Review questions
Question 1
In our calculations of projectile motion, what assumptions are we making?
Question 2
How did Galileo influence the understanding of projectile motion?
Question 3
a
b
c
When a projectile is launched horizontally, which component of its velocity remains
constant?
When a projectile is launched horizontally, which component of its velocity changes
uniformly with time?
If the initial horizontal velocity of the projectile is decreased, will the time taken for
the projectile to fall to the ground be decreased, increased or remain the same?
Explain.
Question 4
Explain the strategy that you would use to calculate a projectile’s:
a
maximum height
b
time in the air
c
velocity
d
range.
Question 5
A coin is flipped upwards with an initial velocity of 3.0 m s1. Calculate the:
a
maximum height that it attains
b
time that it is in the air before it is caught again at the same height from which it
was released.
Question 6
A marble is rolled off a horizontal table of height 1.5 m with an initial speed of
10 m s1. Determine the:
a
time that elapses before the marble hits the ground
b
distance from the table that it travels
c
speed with which the marble hits the ground.
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Question 7
A football is kicked from the ground with a velocity of 25 m s1 at an angle of 60°.
Calculate the:
a
b
length of time that the ball is in the air
the horizontal distance that the football travels.
Question 8
A stone is projected out to sea from the edge of a vertical cliff 125 m high. If the
velocity is 10 m s1 horizontally, find the:
a
b
c
time the stone takes to reach the water
distance from the base of the cliff at which it strikes the water
velocity at that instant.
Question 9
A stone is thrown at a height of 1.4 m with a velocity of 20 m s1 at an angle of 45°
to the horizontal. Determine:
a
b
c
how far it will travel horizontally until it is again at a height of 1.4 m
the time for which it is above the level from which it is thrown
the greatest vertical height reached.
Question 10
A stone is thrown horizontally from a cliff of height 50 m on the Moon’s surface. Given
that the Moon has a gravitational acceleration of 1.67 m s2, calculate the length of time
to reach the bottom of the cliff.
Question 11
A girl throws a ball vertically upwards so that it reaches a height of 20.0 m above her hand.
a
b
c
d
What is the velocity with which the ball leaves the girl’s hand?
Find the time taken to reach the greatest height.
For how long does the ball remain in the air?
For what fraction of the total time is the ball above 15.0 m?
Question 12
Sketch the velocity–time graph and the acceleration–time graph for:
a
b
a freely falling object released from height h
an object projected vertically upwards from the ground.
Question 13
A projectile is launched horizontally at 21 m s1 at a height of 75 cm above the ground.
a
b
c
How long does it take to reach the ground?
Where does it hit the ground?
What is its velocity just before it hits the ground?
Question 14
A shell is fired from a cannon at 350 m s1 at an angle of 40° to the horizontal.
a
b
c
d
e
What is the initial horizontal component of its velocity?
What is the initial vertical component of the shell’s velocity?
What is its horizontal displacement after 5 s?
How high off the ground is it after 5 s?
What is the displacement after 5 s, expressed in terms of its horizontal and
vertical displacement?
Question 15
A bullet is fired from a gun with a horizontal component of velocity of 2000 m s1. The
bullet is to hit a target 6 m above the horizontal and 500 m away. Calculate the:
a
b
time of flight
angle of elevation of the gun.
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