Second Quarter Science Batch Reviewer
By: 10-17 Hope Committee
In collaboration with 10-12 Hope Committee
Lesson I: Projectile Motion
What is projectile motion?
Projectile motion is a kind of motion where an object undergoes a curved path. In other words, a projectile’s
trajectory follows a parabolic motion.
This type of motion occurs in two dimensions: horizontal (x) and vertical (y). A projectile’s motion is under the
influence of gravity.
Trajectory- A trajectory is an arc-shaped path that forms once the projectile is launched or thrown. This encompasses
the comprehensive understanding of how an object moves through space over time, influenced by initial conditions
such as launch angle, initial velocity, and forces acting on the object or projectile.
Projectile- A projectile is any object thrown up into space upon which the only acting force is gravity .
Take note: In the absence of air resistance, the horizontal velocity of a projectile is constant (a never changing in
value). While there is a vertical acceleration caused by gravity at a rate of 9.8 𝑚/𝑠 2 . The horizontal and vertical
motion are independent of each other.
To have a better understanding, observe the photo below:
Horizontal and Vertical Components
X-component (Horizontal motion)
-The X-component neglects air resistance
-There is a constant horizontal velocity (Vx)
-Horizontal acceleration is zero. (ax = 0)
-We call the horizontal distance as range (dx)
Y-component
-The force acting upon this motion is the force of gravity. (g = ay = ag) (g = -9.8 m/𝑠 2 )
-Vertical velocity is not constant (Vy)
-We call the vertical distance as the height (h = dy)
Problem Solving
Key Principles in solving:
-Distances over the point of release are positive while distances below the point of release are negative.
-Upward velovities are positive, downward velocities are negative
-Acceleration due to gravity (g) is always negative.
Sample Problems
Problem 1:
A soccer ball is kicked horizontally with an initial velocity of 30 m/s from the edge of a 2.0-meter-hight platform (a)
How long will it take to reach the ground? (b) How far will it travel horizontally?
Given:
Initial Velocity (Vix): 30 m/s
Acceleration due to gravity (a): -9.8 m/𝑠 2
Vertical displacement (d): -2.0 m/s
Required:
a) t =?
b) dx=?
Equation:
a) dy = Viyt + 1/2g𝑡 2
b) dx = Vxt
Solution:
a) -2.0m = 0 + ½ (-9.8m/𝑠 2 ) 𝑡 2
2.0𝑚 −4.9 𝑚/𝑠 2𝑡 2
=
4.9𝑚/𝑠 2 −4.9𝑚/𝑠 2
Cancel out
0.41 𝑠 2 = 𝑡 2
√0.41𝑠 2 = √𝑡 2
t = 0.64s
b) dx = (30 m/s) (0.64 s)
dx = 19.2 m
Answer:
a) The soccer ball will take 0.64 seconds to hit the ground
b) The soccer ball will travel 19.2 meters horizontally
Problem 2
A marble is launched horizontally from a height of 3.0 meters with an initial velocity of 20 m/s. (a) How long will it
take to reach the ground? (b) How far will it travel horizontally?
Given:
Initial Velocity (Vix) = 20 m/s
Height (dy) = -3.0m
Acceleration due to gravity (a): -9.8 m/𝑠 2
Required:
a) t = ?
b) dx = ?
Equation:
a) dy = Viyt + 1/2g𝑡 2
b) dx = Vxt
Solution:
a) -3.0 = 0 + 1/2 (-9.8m/s2 ) t2
-3.0m / -4.9m/s2 = 4.9m/s2 t2//4.9m/s2
cancel out m/s2
√0.61𝑠 2 = √𝑡 2
t = 0.78s
b) dx = (20m/s)(0.78s)
cancel out “s”
dx = 15.6m
Answer:
a) The marble will take 0.78s (seconds)
b) The marble will travel 15.6m (meters) horizontally
Problem 3
An arrow is fired directly horizontally off a cliff that is 10.0 meters tall, with a velocity of 65.5 m/s.
(a.) How long is the arrow in the air? (b.) What is the range of the arrow?
Given:
Height (h) = 10.0m
Horizontal Velocity (Vx) = 65.5 m/s
g = -9.8 m/s2
Required:
a) Time (t) = ?
b) Horizontal Distance (dx) = ?
Equation:
a) dy = Viyt + 1/2g𝑡 2
b) dx = Vxt
Solution:
a) dy = 1/2g𝑡 2
dy = 4.9
10.0m/4.9 = 4.9 m/s (t2)
√2.04𝑠 = √𝑡 2
1.43s = t
b) dx = (65.5m/s) (1.43s)
cancel out “s”
dx = 93.67m
Answer:
a) was in the air for 1.43s
b) The range of the arrow was 93.67m
Lesson II: Projectile Motion Launched at an Angle
Vertical velocity (Vy)- The vertical velocity is responsible for the projectile to travel at a vertical distance.
As a projectile rises, the vertical velocity starts to decrease, reaching zero at the top of its point. Although
as it starts to drop, the vertical velocity increases due to acceleration caused by gravity.
Horizontal velocity (Vx)- The horizontal velocity is responsible for the projectile to travel at a horizontal
distance. The horizontal distance is always constant
Angle- An angle is a numerical value measured in degrees that determines the orientation of the projectile
to be thrown. The angle determines the range, height, and time of flight it will experience when in
projectile motion.
You may refer to the image below:
As you can observe, the higher the angle, the higher the height the projectile will travel, in this case would
be 75. Meanwhile, to achieve the maximum range, you must achieve an angle of 45 degrees.
Take note: A projectile launched either at 30 degrees or 60 degrees will have the same displacement.
Although launched together, the shorter angle will have a shorter time of flight
Problem Solving
Below are the equations:
Key Principles:
Step 1: Calculate the initial velocity for both horizontal and vertical components (Vix, Viy)
Step 2: Calculate the time of flight
Step 3: Calculate the total time of flight (tT)
Step 4: Calculate the height (dymax)
Sample Problem 1
A projectile is launched at 32.1 m/s and 52.6° above the horizontal. Determine the time in the air, the
horizontal displacement, and the peak height.
Given:
Initial velocity (Vi) = 32.1 m/s
Angle of projection θ = 52.6°
Required:
dy max = ?
dx = ?
t= ?
Equation:
Step 1: Calculate the value of Vix and Viy
Vix=Vi.cosθ
Vix = (32.1m/s) . cos(52.6)
Viy = Vi . sinθ
Viy = (32.1m/s) . sin(52.6)
Vix = 19.50 m/s
Step 2: Calculate the height (dy max)
(𝑉𝑦)2−(𝑉𝑖𝑦)2
dy max = 2𝑔
(0)2−(25.50 𝑚/𝑠)2
dy max = 2(−9.8 𝑚/𝑠2)
−650.25 𝑚2 𝑠 2
dy max = −19.6 𝑚/𝑠2
dy max = 33.18m
Step 3: Calculate the horizontal displacement (dx)
dx = Vix . tT
𝑉𝑦−𝑉𝑖𝑦
t= 𝑔
0−25.50 𝑚/𝑠
t= −9.8 𝑚/𝑠 2
−25.50 𝑚/𝑠
t= −9.8 𝑚/𝑠 2
t = -2.60 s
Calculate the time (tT):
tT = 2(t)
tT =2(2.60s)
dx = Vix . tT
dx = 19. 50 m/s . 5.20 s
dx = 101.40 m
