Initial Velocity and Launch Angle

Expert reviewed • 22 November 2024 • 8 minute read

What is Initial Velocity?

The initial velocity $(u)$ is the speed and direction at which an object is launched. It’s a vector quantity, meaning it has both magnitude and direction. In projectile motion, this initial velocity can be broken down into two components:

Horizontal component $(u_x)$
Vertical component $(u_y)$

Vector Resolution

To analyse projectile motion effectively, we resolve the initial velocity vector into its horizontal and vertical components using trigonometry. This process involves creating a right-angled triangle where:

The hypotenuse represents the initial velocity $(u)$
The adjacent side represents the horizontal component $(u_x)$
The opposite side represents the vertical component $(u_y)$
The angle between the hypotenuse and the adjacent side is the launch angle $(\theta)$

Mathematical Relationships

Using trigonometric ratios, we can express the horizontal and vertical components of velocity in terms of the initial velocity and launch angle:

Horizontal component: $u_x=ucos θ$
Vertical component: $u_y=usin θ$

The relationship between these components can also be expressed as:

\frac{u_y}{u_x} = \tan \theta

Additionally, we can use Pythagoras’ theorem to relate the magnitude of the initial velocity to its components:

u^2 = u_x^2 + u_y^2\\\text{or}\\u = \sqrt{u_x^2 + u_y^2}

What is Launch Angle?

The launch angle $(\theta)$ is the angle at which the projectile is launched relative to the horizontal. It plays a crucial role in determining the trajectory of the projectile.

Effect on Components

A larger launch angle increases the vertical component $(u_y)$ and decreases the horizontal component $(u_x)$ .
A smaller launch angle does the opposite, increasing $u_x$ and decreasing $u_y$ .

Special Cases

At $0\degree$ , the projectile is launched horizontally $(u_y = 0, u_x = u)$
At $90\degree$ , the projectile is launched vertically upward $(u_x = 0, u_y = u)$

Practice Question 1

A ball is launched with an initial velocity of 50 m/s at an angle of 37° above the horizontal. Calculate its initial horizontal and vertical velocity components.

Initial horizontal velocity:

u_x = ucos θ \\= 50cos 37° \\= 39.9\: m/s

Initial vertical velocity:

u_y = usin θ \\= 50sin 37° \\= 30.1\:m/s

Changes in Velocity During Projectile Motion

Understanding how velocity changes during projectile motion is crucial:

Horizontal velocity $(u_x)$ : Remains constant throughout the motion (assuming no air resistance).
Vertical velocity $(u_y)$ : Changes continuously due to gravity:
- Decreases when the projectile moves upward
- Increases when the projectile moves downward
- Reaches zero at the highest point of the trajectory

What are the Equations of Motion for Vertical Components?

To analyse the vertical motion of a projectile, we use the equations of motion for uniformly accelerated motion:

s = u_y t + \frac{1}{2}a_y t^2

v_y = u_y + a_yt

v_y = u_y + 2a_ys

Where:

$s$ is the vertical displacement
$a_y$ is the acceleration in the vertical direction (gravity, -9.8 m/s²)
$u_y$ is the initial vertical velocity
$v_y$ is the vertical velocity after time $t$

Practice Question 2

A projectile is launched with an initial velocity of 60 m/s at an angle of 45° above the horizontal.

Calculate:

The initial vertical and horizontal velocity components
The vertical velocity after 2 seconds
The instantaneous velocity (magnitude and direction) after 2 seconds

Initial velocity components:

u_x = 60cos 45° \\= 42.4 m/s

u_y = 60sin 45° \\= 42.4 m/s

Vertical velocity after 2 seconds:

v_y = u_y + a_yt \\= 42.4 + (−9.8)(2) \\= 22.8\:m/s

Instantaneous velocity after 2 seconds:

Magnitude:
$v = \sqrt{v_x^2 + v_y^2} \\= \sqrt{42.4^2 + 22.8^2} \\= 48.1\:m/s$
Direction:
$\theta = \tan^{-1}(\frac{v_y}{v_x}) \\= \tan^{-1}(\frac{22.8}{42.4}) \\= 28.3°$

Understanding Projectile Motion

Up Next

Maximum Height, Time of Flight, and Range

Return to Module 5: Advanced Mechanics