Projectile Motion

• The motion occurs under the influence of gravity, which acts downward throughout.
• The path followed by the object is a parabolic curve.
• The motion can be divided into two independent components:

Horizontal motion: In horizontal motion, the object moves at a constant velocity because there is no horizontal acceleration.

Vertical motion: In vertical motion, the object experiences acceleration due to gravity.

• The horizontal and vertical motions are independent of each other but happen simultaneously.
• At the highest point of the trajectory, the vertical velocity is zero, but the horizontal velocity remains constant.

Projectile motion equations describe the object’s motion in terms of its horizontal and vertical components. 

Horizontal Motion

In horizontal motion, the object moves at a constant velocity because there is no horizontal acceleration. The horizontal distance (x) traveled can be calculated using:

x = v_x t  

Where 

– v_x: Horizontal velocity

– t: Time 

Vertical Motion

In vertical motion, the object experiences acceleration due to gravity (g = 9.8 m/s²). The vertical position (y) is given by:

 y = v_y0 t − ½ g t²

The vertical velocity at any time is:

v_y = v_y0 − g t

Where:

– v_y0​: Initial vertical velocity.

The following table summarizes the key terms associated with projectile motion. The formulas are derived from the above equations of motion.

Term	Definition	Symbol and Formula
Launch Angle or Angle of projection	The angle at which the projectile is launched relative to the horizontal axis.	θNot directly calculated
Initial Velocity	The speed at which the projectile is launched, determining the horizontal and vertical components of the motion.	v0Not directly calculated, but its components are as follows:Horizontal: vx0 ​= v0 ​cos θVertical: vy0 = v0 sin⁡ θ
Time of Flight	The total time a projectile remains in the air from launch to landing.	T = 2 vo​ sin θ/ g
Horizontal Range	The horizontal distance traveled by a projectile before hitting the ground.	R = v02 ​sin 2θ/g​
Maximum Height	The highest vertical point is reached by the projectile during its motion.	H = v02​ sin2 θ​/ 2 g

• Sports: Predicting the trajectory of a football, basketball, or golf ball to improve performance and strategy.
• Ballistics: Determining the path of bullets, missiles, or artillery shells for military and defense purposes.
• Engineering: Designing structures like parabolic bridges, roller coasters, and water fountains that rely on projectile motion principles.
• Space Science: Calculating the launch angles and velocities for rockets, satellites, and interplanetary probes.

Problem 1: A ball is thrown horizontally from a height of 20 m with an initial velocity of 10 m/s. Find the time it takes to hit the ground and its horizontal range. (Acceleration due to gravity, g = 9.8 m/s²)

Solution

Given:

Height, h = 20 m

Initial velocity, v₀ = 10 m/s

Angle of launch, θ = 0° (since it is thrown horizontally)

Therefore, the initial horizontal velocity is:

v_x0 ​= v₀ ​cos θ =  v₀ ​cos 0° = v₀

The initial vertical velocity is:

v_y0 = v₀ sin⁡ 0° = 0

Since the ball is thrown horizontally, only the vertical motion is affected by gravity. The vertical motion follows the equation:

\[ y = v_{y0} t + \frac{1}{2} g t^2 \]

Since v_y0 = 0, the equation simplifies to:

 \[ y = \frac{1}{2} g t^2 \]

We can solve for time t when the ball hits the ground (y = 20 m):

\[ 20 = \frac{1}{2} (9.8) t^2 \\

=> t^2 = \frac{20 \times 2}{9.8} = \frac{40}{9.8} \approx 4.08\\

=> t \approx \sqrt{4.08} \approx 2.02 \text{ seconds} \]

Therefore, it takes 2.02 seconds to reach the ground.

The horizontal range depends on the horizontal velocity and the time of flight. Since there is no horizontal acceleration, the horizontal distance is given by:

R = v_x t = 10 m/s x 2.02 s = 20.2 m

The horizontal distance it travels is 20.2 m.

Problem 2: A projectile is launched with an initial velocity of 30 m/s at an angle of 45°. Calculate its time of flight, maximum height, and range. (g = 9.8 m/s²)

Solution

Given:

Initial velocity, v₀ = 30 m/s

Launch angle, θ = 45°

The time of flight is calculated from the following formula:

T = 2 v_y0​/ g

Where:

v_y0 = v₀ sin⁡ θ = 30 m/s x sin⁡ 45° = 21.2 m/s

Therefore,

T = 2 x 21.2 m/s ​/ 9.8 m/s² = 4.33 seconds

The total time the projectile stays in the air is 4.33 seconds.

The maximum height H is given by:

H = v_y0²​​/ 2 g = (21.2 m/s)²/ 2 x 9.81 m/s² = 22.9 m

The maximum height it reaches is 22.9 m.

The range R is given by:

R = v₀² ​sin 2θ/g​ = (30 m/s)² x sin (2 x 45°)/ 9.81 m/s² = 91.7 m

The horizontal distance the projectile travels is 91.7 m.