SCIENCE
Quarter 4 – Module 1
Projectile motion
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Science – Grade 9
Quarter 4 – Module 1
MELC: 1. Describe the horizontal and vertical motions of a projectile (S9FE-Iva-34)
2. Investigate the relationship between the angle of release and the height and range
of the projectile (S9FE-Iva-35)
After this module, you are expected to learn the following:
1. Identify the difference between Uniform Motion and Uniform Accelerated Motion.
2. Describe the Uniform Accelerated Motion along Horizontal and Vertical dimensions
qualitatively and quantitatively.
3. Define projectile motion
4. Describe the horizontal and vertical motions of a projectile (S9FE-Iva-34)
5. Apply kinematic equations and vectors to solve problems involving projectile motion.
6. Investigate the relationship between the angle of release and the height and range of the
projectile (S9FE-Iva-35)
LESSON 1: Projectile Motion
The KEY
CONCEPTS

Vertical and Horizontal motions of a projectile are independent of
each other.
Air resistance- is a force that acts in the opposite direction to an object
moving through the air.
Free-fall – is the motion of an object where gravity is the only force acting upon it.
Maximum Height- the maximum vertical displacement travelled by the projectile in its
trajectory.
Non-uniform Accelerated Motion- is a motion in which the change in velocity with time in not
linear or uniform.
Projectile- the object being thrown (projected) into the air
Projectile Motion- movement experienced by a body thrown at an angle not equal to 90 degrees
and follows a curvilinear path.
Range- the horizontal distance travelled by a projectile.
Time of Flight- the entire duration while the projectile is in its trajectory.
Trajectory- the path that a projectile follows.
Uniform Accelerated Motion- is a motion in which the change in velocity with time is linear or
uniform.
Uniform Motion- is a kind of motion in which an object travels equal distance in equal time
intervals or at constant velocity.
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Review
What is Motion in a Straight Line?
If an object changes its position with respect to its surroundings with time, then it is
called in motion. It is a change in the position of an object over time. Motion in a straight line is a
linear motion or motion along a straight line.
The linear motion is also called the Rectilinear Motion can be of two types:
1. Uniform linear motion with constant velocity or zero acceleration.
2. Non-uniform linear motion with changing velocity or non-zero acceleration.
What is the difference between Uniform Motion and Non-Uniform Motion?
The simplest type of motion is uniform motion. For uniform motion, the velocity is
constant (v=k), the acceleration is zero (a=0), and the instantaneous velocity is equal to the
average velocity. Its displacement dx may be obtained by multiplying its constant velocity v by the
time t. Example, the speed of car is 10 m/s.
dx= vt
Let’s solve!!
A car moving at constant speed travels 30 m in 5 s.
(a) What is the speed of the car? (b) How far will the object move in 10 s?
Given:
d = 30 m
t= 5 s
Find: v and dx at 10s
Formula: dx= vt
v = dx/t
Solve:
(a) v = dx/t
v = 30 m/5 s
v = 6 m/s
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(b) dx= vt
dx= 6 m/s (10 s )
dx= 60 m
Try this….
1. A train leaves Station A at a constant speed and arrives at Station B in 8 hours. A
second train leaves Station A at a constant rate of 40 km/h and arrives at Station B
in 10 hours. What was the constant speed of the first train?
The Non-uniform motion is a motion of an object in which the object travels with varied
speed and it does not cover the same distance in equal time intervals, irrespective of the time interval
duration. The acceleration is non-zero. Example, a car covers 10 meters in first two seconds, and
15 meters in next two seconds.
An important special case of non-uniform motion along a line is that which arises
when an object is subjected to constant acceleration. This kind of motion is called uniformly
accelerated motion .
Uniformly Accelerated Motion-a body will move with a constant acceleration (a=k) and its velocity
will keep on changing with time at a constant/steady rate (v=changing).
In general, a uniform accelerated motion is the one in which the acceleration of the object
throughout the motion is uniform. Examples, a ball rolling down a slope, a skydiver jumping out of a
plane, and a ball dropped from the top of a ladder.
Here are the equations used in uniform accelerated motion. Most often, these equations
are used to describe either horizontal or vertical motion.
Remember, acceleration must be constant.
d=Vf +Vi t
2
d = V i t + ½ at2
V f= V i + at
V f 2 = Vi2 + 2 ad
Where,
Vf = is the final velocity
Vi =is the initial/original velocity
a = acceleration
d=displacement
t= time
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Uniformly Accelerated Motion: Horizontal Dimension
If an object maintains a constant change in its velocity in a given time interval along a
straight line, then the object is said to have a uniform acceleration.
dx = V f + V i t
2
V f= V i + at
dx = V i t + ½ at2
V f 2 = Vi2 + 2 adx
Let’s solve!!!
An airplane from rest accelerates on a runway at 5.50 m/s 2 for 20.25 s until it finally takes
off the ground. What is the distance covered before takeoff?
Given:
vi = o m/s (at rest)
a = 5.5 0 m/s2
t = 20.25 s
Find: d
Formula: d = vit + ½ at2
Solution:
0
d= vit + ½ at2
d= (0 m/s)(20.25 s ) + ½ ( 5.50 m/s2 )(20.25 s)2
d= ½ (5.50 m/s2 )(410.06 s2)
d= ½ (2,255.33 m )
d=1,127.67 m
Try this…
a. A jeepney from rest accelerates uniformly over a time of 3.25 seconds and covers a
distance of 15 meters. Determine the acceleration of the jeepney.
Uniformly Accelerated Motion: Vertical Dimension- Free Fall
Perhaps you have observed that what goes up always comes down. In the absence of
air resistance, it is found that all bodies regardless of size and weight at the same location above
the earth’s surface fall vertically with the same acceleration. This very idealized motion in which air
resistance is neglected and the acceleration is constant is called free fall. Free Fall is a state where
an object moves under the sole influence of gravity. The acceleration of a freely falling body is
called acceleration due to gravity. This is denoted by a letter “g” and is equal to -9.8 m/s2 at the
surface of the earth. This acceleration is downward and directed toward the center of the earth.
We may treat free fall as a case of uniformly accelerated motion. All the equations that
we have for uniformly accelerated motion may be used to analyse free-fall. Since motion is along
the vertical direction, it is better to replace dx with dy in our equations and a with g.
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Correspondingly we have,
dy=Vf +Vi t
2
V f= V i + gt
d y = V i t + ½ gt2
V f 2 = Vi2 + 2 gdy
The following sign conventions shall also be used:
a. Distances above the origin are positive, while distances below the origin are
negative.
b. Upward velocities are positive, downward velocities are negative.
c. g is always negative.
 Note. The origin is the point of release.
Let us now have our examples and see when to use each equation. In free-fall, there are
three possible situations:
CASE 1: Dropping (vi=0)
This is the simplest case- in dropping the ball, the initial velocity is zero. The acceleration
due to gravity throughout the flight is 9.8 m/s2 downward or -9.8 m/s2.
Let’s solve!!
A ball is dropped from a hill top. If it takes 5 s for the stone to hit the ground, how high is the
hill? What is the velocity just before it hits the ground?
Given: vi=0 since the object is dropped.
t = 5s
g = 9.8 m/s2 downward
Find: (a) dy
(b) vf
Solution:
0
(a) dy= vit + ½ gt2
dy= 0 + ½ (9.8 m/s2 )(5 s)2
dy= ½ (9.8 m/s2 )(25 s2)
dy= 1/2 (245 m)
dy= 122.5 m
0
(b) vf= vi+gt
vf= 0+(9.8 m/s2 )(5 s)
vf= 49m/s, downward
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Note: The negative sign in the acceleration due to gravity “g” indicates direction. Since we are
using dy which indicates the direction of the object- downward, so we are using g as 9.8 m
CASE 2: Thrown Downward (vi, downward)
This is actually the general case of free fall- the object is given an initial velocity.
Let’s Solve!!
From the top of a 122.5 m hill, a stone is thrown
downward with a velocity of 10 m/s. What is the
velocity just before it hits the ground? How long does
it take the stone to reach the ground?
Given: vi= 10 m/s, downward
dy = 122.5
Find: vf and t
Formula: (a) Vf 2 = vi2+2gd
(b) Vf = vi + gt
Solution:
(a) Vf 2 = vi2+2gd
Vf = √vi2+2gd
Vf = √(10 m/s)2 + 2 (9.8 m/s2 )(122.5 m )
Vf = √100 m2/s2 + 2,401 m2/s2
Vf = √2,501 m2/s2
Vf = 50 m/s, downward
(b) Vf = vi + gt, rearrange the formula to find
for t. Then,
t = vf - vi
g
t = 50 m/s-10 m/s
9.8 m/s2
t = 40 m/s
9.8 m/s2
t = 4.08 s
CASE 3: Thrown Upward (vi , upward)
In this case we would have two types of motion, upward and then downward. We would
be treating each case differently.
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When an object is thrown upward, its displacement and velocity is upward, but
acceleration, g acts downward, causing the object to slow down until it stops. Once the object stops,
its behaviour is similar to case 1, but the height is higher because of the initial upward motion.
The best way to show this type of motion is through this example.
Take note, at ALL TIMES during the
flight, the acceleration is ALWAYS equal
to 9.8 m/s2 downward, even if the ball
stops at the highest point.
Sample Problem:
The height of the boy is 1.23 m, he throw a stone upward with a velocity of 10 m/s. What is the
velocity just before it hits the ground? How long does it take the stone to reach the ground?
Given:
h = 1.23 m
vi = 10 m/s
Find:
Final velocity and time for the stone to hit the ground.
Solution:
First we must solve for the new height of the boy due to the upward velocity. In this
condition, ALWAYS, the final velocity is ZERO. So, we have added a new given, V f =0. Solving for
the additional height d upward, we have a given of vi, vf, and of course the g.
So, use equation: vf 2 = vi2 + 2 gd
,
let d = d up
d up = vf 2 - vi2
2g
Note: The direction is upward, and g is downward, so in this upward motion, use
g = - 9.8 m/s2
d up = vf
2
- vi2
d up = 0 – (10 m/s
2 (- 9.8 m/s2 )
2g
d up = - 100 m2/s2
-19.6 m/s2
d up = 5.1 m
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dtotal = d up + d boy
d total = 5.1 m + 1.23 m
d total = 6.33 m
Now that we know the new height, let us call it d total . We can treat the rest as simple free fall from
maximum height reach by the object. The initial velocity is ZERO.
a. Vf2 = vi2 + 2 gd, since vi is zero
vf =√2 𝑔𝑑
Vf = √2 (9.8m/s2) (6.33 m)
Vf = √2 (62. 03 m2/s2 )
Vf = √124.06 m2/s2
Vf = 11.14 m/s
b. Vf= vi+gt, since vi is zero because the object is at the maximum height.
Vf = gt
t = v f /g
t = 11.14 m/s
9.8m/s2
t = 1.14 s
Remember:
The motion of an object thrown upward and eventually returning to its starting point exhibits
two symmetries:
a. Time symmetry- means that the time required for the object to reach maximum height
equals the time for it to return from its maximum height to its starting point.
b. Speed symmetry- shows that at any displacement above the point of release, the
speed of the body during the upward trip equals the speed during the downward trip.
WHAT IS A PROJECTILE MOTION?
We see one dimensional motion in previous topics. Now, we will try to explain motion in
two dimensions that is exactly called “projectile motion”. In this type of motion gravity is the only
factor acting on our objects.
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The motion of a body thrown horizontally or at an angle ø other than 90 degrees with the
horizontal is called PROJECTILE MOTION.
We can have different types of projectile type. For example, you throw the ball straight
upward, or you kick a ball and give it a speed at an angle to the horizontal or you just drop things
and make them free fall; all these are examples of projectile motion.
What are the factors that affect a projectile motion?
Factors affecting the flight path of a projectile are:
 Gravity
 Air resistance
 Speed of release
 Angle of release
 Height of release
 spin
A projectile is an object upon which the only force acting is gravity. The path that a
projectile follows is called its trajectory. The trajectory of a projectile is a parabola. This results from
the combined motions in the horizontal and vertical directions.
 Note that in projectile motion, gravity is the only force acting on the object.
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 The kinematic equations of motion that we previously discussed applies to projectile
motion. Remember that projectile motion has two components: horizontal
component and vertical component. The horizontal and vertical motion of a
projectile are independent, meaning they do not affect each other. Since vertical
and horizontal motions are independent, we can analyze them separately.
Slideplayer.com
PROJECTILE LAUNCHED HORIZONTALLY
In the picture given above, we have a projectile motion and velocity components at
different positions. At the beginning the ball is thrown with an angle to the horizontal. V is its velocity
and its direction is northeast. Vx and Vy are the X and Y components of our velocity. If we look at
the forces acting on our ball we see only the gravity as a force.
In vertical, you can see in the picture that the velocity (vy) decreases as the object rises.
As it reaches its maximum height, vertical component of our velocity (vy) becomes zero but its
acceleration at the highest point is 9.8 m/s2 and not zero as in the case of free fall examples. After
Vy becomes zero our ball changes its direction and make free fall now. As the object falls towards
the Earth again, the vertical velocity (vy) increases again in magnitude but points in the opposite
direction to the initial vertical velocity. You will see that it has a “-“sign in front of the Vy because its
direction is downward. Finally, when the ball hits the ground Vy reaches its beginning magnitude but
opposite in direction. We see the effect of gravity on vertical motion.
Now, let’s examine the horizontal part of our projectile motion. This part is so easy that
you can understand from the picture, our horizontal component of velocity is constant during the
motion. Why it is constant? What changes the velocity?
In previous lesson we learned force concept that causes change in the state of motion.
Look at our horizontal motion carefully. Is there any force acting on our object in horizontal
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direction +X or –X? The answer is actually no. However, in –Y direction gravity is acting on our
object which makes Vy decrease and becomes zero at the top.
The x- and y- motions are recombined to give the total velocity at any given point in the
trajectory.
Displacement
Velocity
d= √ x2 + y2
v= √vx2 + vy2
ø= tan-1 (y/x)
ø= tan-1 (vy/vx)
All these explanations say that, we have two motions in projectile motion. One of them is
constant motion in horizontal and other one is free fall under the effect of gravity in vertical. We tried
to explain projectile motion with words. Now it is time to give equations of motion under two titles.
Let us have this Table1 that summarizes the equations applicable for each component of
projectile motion. Vix refers to the initial horizontal velocity. Viy refers to the initial vertical velocity.
Table 1: THE HORIZONTAL AND THE VERTICAL COMPONENTS OF PROJECTILE
MOTION
HORIZONTAL
VERTICAL
Acceleration
ax=0
ay=g
Velocity
Vix=vi cos ø
Viy= vi sin ø
Vx=vix
Vy= viy + gt
dx= vxt
dy=viy t + ½ gt2
Displacement
2gdy= vy2 – viy2
For a projectile fired at an angle ø above the horizontal, the projectile rises to some maximum
height and then descends. It finally lands at some horizontal distance from its launching point. This
horizontal distance is called Range. The range is the horizontal distance travelled by the projectile
between the launching position and the landing position, on the assumption that it will return to the
same level at which it is fired. Range depends not only on the projection speed but also on the angle
of projection. Maximum range is obtained at an angle of projection of 450.
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PROJECTILE LAUNCHED AT AN ANGLE
ELEMENTS OF A PROJECTILE MOTION
Where,
Vo is the velocity of an object
Ø is the angle
H is the maximum height
R is the range
Vx is the velocity of a projectile on horizontal motion
Vy is the velocity of a projectile on vertical motion
Projectile motion also exhibits time and speed symmetries. In addition, there are always
two angles giving the same range for a given initial velocity. These two angles are complementary.
Complementary angles because they add up to 90°. Examples, angles of projection of 150 and
750 ; 600 and 300, for a given initial velocity cover the same range.
Slideplayer.com
The figure above shows the paths of several projectiles all having the same initial speed
but different projection angles. Neglecting the effects of air resistance the paths are all parabolas.
Notice that these projectiles reach different heights (altitude) above the ground. They also travel
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different horizontal distances,that is, they have different horizontal ranges.As the angle increases,
the height also increases. If the angle is less than 45 degrees, the range is increasing. If the angle
is greater than 45 degrees the range is decreasing. As the angle increases, the time also
increases.
Sample problem:
A ball is thrown horizontally from a height of 5.50 m with an initial speed of 25.0 m/s.
A.
B.
C.
D.
How long will it take the ball to reach the ground?
At what horizontal distance from the point of release will it strike the ground?
What will be the magnitude of its velocity when it strikes the ground?
At what direction will it strike the ground?
Solution:
This is an example of a projectile where the angle of projection is equal to zero. We are
given that initial velocity vi=25.0 m/s, and height dy= 5.50 m. We resolve the initial velocity into its
horizontal and vertical components,
Vix= 25.0 m/s cos 0 = 25.0 m/s
Viy= 25.0 m/s sin 0 = 0
A. We consider the point of release as the origin. Taking into account only the vertical
component of motion, dy=-5.50 m. Substituting the given values in the equation
dy=viyt+1/2 gt2 = ½ gt2
t= √2dy/g
t= 1.06 s
B. Consider the time the horizontal component of motion
dx= vix t
dx= 25.0 m/s (1.06 s )
dx= 26.5 m
C. The magnitude of its velocity is given by V= √vx2+vy2
We know that vx=vix=25.0 m/s
We first solve for the vertical component of its velocity.
Vy= viy + gt
Vy= 0 + (-9.8 m/s2 )(1.06 s )
Vy= -10.39 m/s
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Substituting the values we obtained and solving for v
V= √ (25.0 m/s)2 + (-10.39 m/s )2
V=√625 m2/s2 + 107.95 m2/s2
V=√732.95 m2/s2
V= - 27.07 m/s, because the ball is going down.
D. The direction at which it will strike the ground is
Ø= tan-1 Ivy/vxI
Ø= tan-1 -10.39 m/s
25.0 m/s
0
Ø= 22.6 , north of east
Special case:
Equations:

tf= -2 vi sin ø t
g

dy= -vi2 sin2 ø
2g

R= -vi2(sin 2ø)
g
Note: sin 2ø =1
Sample problem:
The world record in the javelin throw is 91.46 m. If you throw this at an angle of 45 degrees,
what should be your initial velocity? Use the equation for the special case.
Given: R = 91.46 m
ø=45 0
Find:
Initial velocity, vi
Solution:

From, R= -vi2(sin 2ø)
vi= √ -Rg
g
Vi=√-
sin 2ø
91.46 m (-9.8 m/s2)
Sin 2(450)
Vi= √896.308 m2/s2
Vi= 29.94 m/s
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References:
1. Science 9 Learner’s Module
2. Exploring Life Through Science- Physics. Silverio, Angelina A. (2007) p.47-61
3. E-Physics, Santos, Gil Nonato S. and Ocampo, Jorge P. (2003) p. 155-202
4. Conceptual Physics by Paul Hewitt (7th Edition)
5. https://openstax.org.physics
6. https://www.saratogaschools.org
7. https://www.khanacademy.org
8. https://www. bbc.co.uk
9. https://courses.lumenlearning .com
10. https://www. basd.net.Physics
11. https://www. pinterest.com
12. https://www.physicsclassroom.com
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