PROJECTILE MOTION
2.4
NCEA LEVEL 2 PHYSICS
CONTENTS
 Introduction.
 Gravity.
 Projectile Motion.
 Projectile motion calculation summaries.
 Projectile motion at an angle.
 Video “Pool Projectile Motion”.
 Exercises from Rutter.
 Homework
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INTRODUCTION
All projectile motion is caused by the effect of a force. This force we
know is GRAVITY.
Gravity was first understood and explained by Sir Isaac Newton
(1642-1727).
Newton proposed that gravity is
produced due to the interaction between
object (more on that in Level 3). The
effect of gravity on an object that is
moving is that it causes a parabolic
trajectory.
Gravity is unlike other forces as it is
always attractive. All other forces can be
either attractive or repulsive.
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Acceleration due to gravity is constant over the surface of the earth
and a few kilometres into the atmosphere. We know it as 9.81ms-2 or
rounded to 10ms-2.
An object dropped from a height will accelerate straight down at
10ms-2, until it reaches terminal velocity (which we discussed in the
motion topic).
Using kinematic equations of motion we are able to solve many
problems involving gravity, when an object is dropped.
vi is always 0
a = 9.81ms-2
d = vit +
½at2
d = ½(vi + vf)t
vf2 = vi2 + 2ad
Using these
formula d, t, &
vf can be
calculated
vf = vi + at
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Example 1:
A rock (initially stationary) is
dropped from a cliff 50m above
the sea.
a. How far does it fall in 1.0s?
b. How far does it fall in 2.0s?
c. How long does it take to fall
50m?
d = 50m
g = 10ms-1
v/t graph
SOLUTION:
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v(ms-1)
For a. and b., a velocity time graph
can be used. An acceleration of
10ms-2 downwards means that the
rock’s velocity increases by 10ms-1
each second.
Slope 10ms-2
2
1
0
1.0 2.0
3.0
t(s)
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a. For 1.0s: (area under graph = distance)
d = ½ x 1.0 x 10
d = 5.0m
b. For 2.0s:
d = ½ x 2.0 x 20
d = 20m
c. A kinematic equation canbe used to find the time, t. Since the
acceleration of the rock is a = 10ms-2 & vi = 0ms-1 (the rock is
stationary), these values can be substituted into the equation:
d = vit + ½at2
50 = o x t + ½ x 10 x t2
t2 = 50/5.0
t = 10
t = 3.2s
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Example 2:
A ball is thrown upwards with
an initial speed of 20ms-1
a. Draw a velocity time graph
of the motion.
b. How far does it rise in 1.0s?
c. How long does it take to
reach its highest point in its
motion?
d. How high above the ground
is the highest point in its
motion?
vi = 20ms-1
a = 10ms-1
d/t =?
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SOLUTION:
b. Height risen in 1.0s =
area under the graph.
Slope -10ms-2
20
vupwards (ms-1)
a. The velocity/time
graph for the
motion is shown
alongside.
10
0
1
2
3
4
t(s)
-10
-20
area = ½ x (20 + 10) x 1.0
area = 15m
c. At the highest point, the
ball is momentarily
stationary (and is about
to move downwards). At
this moment v = 0 and
occurs at t = 2.0s.
d. Highest position at t =
2.0s. This height is the
area under the graph up to
2.0s.
Area = ½ x 20 x 2.0
 Height = 20m
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READ INFORMATION PAGE 32
COMPLETE QUESTIONS
FROM
RUTTER
9
PROJECTILE MOTION
Any object that moves through the air without its own source of
power. E.g bullets, shot puts, netball, water jets and softballs.
Projectile motion has two major components:
Y component
Resultant
X component
Y component + X component = Resultant
1.
2.
Vertical motion (y component), changes due to gravitational
acceleration, changing at 9.8 ms-2.
Horizontal motion (x component), remains constant as
horizontally it is not subjected to gravity.
N.B: Air resistance is ignored at this level.
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Vertical
component:
Gravity acts
upon the
object
changing the
velocity at
9.8ms-2.
Horizontal
component:
No gravity
acts upon the
object. Thus
the object
moves at a
constant
velocity.
Result of the horizontal & vertical
vectors is a parabolic trajectory
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PROJECTILE MOTION CALCULATIONS
1.
Draw a diagram to represent all the information that you
have been given.
i.
Draw the projectile motion arc
Draw the resultant or y/x component information.
Place on the vf = 0, vertical value.
Time lines.
Range line, calculated using horizontal values using ONLY
v = d/t.
ii.
iii.
iv.
v.
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2.
i.
If you have the resultant velocity i.e. the speed and the angle
then calculate the
Y component:
hyp
y
adj
initial y component = cos θ adj/hyp
ii.
X component:
hyp
opp
x
initial x component = sin θ opp/hyp
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3.
Finding the maximum height of flight:
All of these have to be resolved vertically.
i.
Find the y component velocity, vi
ii.
At maximum height the velocity is 0.  vf = 0.
iii.
Acceleration is constant so a = 9.8 ms-2.
iv.
Distance is unknown,  d = ?.
v.
Use the equation vf = vi + 2ad.
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4.
Finding the time of flight:
i.
Resolve vertically.
ii.
You know vi (y component)
iii.
You know vf = 0 as final velocity.
iv.
You know a = 9.8 ms-2.
v.
Use vf = vi + at.
vi.
Rearrange to find ‘t’. This is the time for half of the flight.
vii.
To find total flight time multiply ‘t’ by 2.
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5.
Finding the range:
i.
Need to resolve horizontally using ‘v = d / t’
ii.
You need to have the full time of flight ‘t’.
iii.
You need to find the initial horizontal velocity ‘x component’.
iv.
Rearrange above formula to make ‘d’ the subject.
v.
Solve.
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Example 3:
A soccer player kicks a ball. The initial components of the velocity
of the ball are 10ms-1 vertically and 20ms-1 horizontally. Find the
range of the ball.
Max height vf = 0
10ms-1
20ms-1
Range = ?
Time = ?
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SOLUTION:
At maximum height, the initial vertical speed of 10ms-1 will have
reduced to zero:
vf = vi + at
0 = 10 + (-10 x t)
t = 1.0s
Time of flight = twice the time to reach max height
T = 1.0 x 2
= 2.0s
The ball travels horizontally at a constant speed of 20ms-1 for 2.0s.
Thus it has a range of:
d = vt
= 20 x 2.0
 range = 40m
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Example 4:
How high does a softball rise if it is hit with an initial speed of 40ms1 at an angle of 40o to the ground?
Max height vf = 0
40o
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SOLUTION:
Initial vertical speed = 40sin40o
= 25.7ms-1
At the maximum height, the vertical speed will have reduced from
25.7ms-1 to zero.
The relevant kinematic equation that does not involve time is:
vf2 = vi2 + 2ad
0 = 25.72 + (2 x -10 x d)
[rearrange]
d = -(25.7)2/ -20
d = 33m
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READ INFORMATION PAGE 33 - 36
COMPLETE QUESTIONS
FROM
RUTTER
COMPLETE HOEWORK
WORKSHEET ON
PROJECTILE MOTION
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