Optimal Angle for Maximum Projectile Range

Expert reviewed • 22 November 2024 • 5 minute read

Review of the Mathematical Foundation of Projectile Motion

To analyse projectile motion, we break down the initial velocity into its horizontal and vertical components:

Horizontal component:

v_x = v_0cos θ

Vertical component:

v_y=v_0sin\theta

Where,

$v_0$ is the initial velocity
$\theta$ is the launch angle.

Deriving the Range Equation

The range (R) of a projectile is the horizontal distance it travels before returning to its initial height. We can derive the range equation as follows:

Time of flight (t):

t = \frac{2v_y}{g} \\= \frac{2v \sin \theta}{g}

Horizontal distance:

R = v_x t \\= (v \cos \theta) \cdot (\frac{2v \sin \theta}{g})

Simplifying:

R = \frac{v^2 \sin 2\theta}{g}

This equation is crucial for understanding why 45° yields the maximum range.

Why Does 45° Maximise Range?

The range equation $R = \frac{v^2 \sin 2\theta}{g}$ shows that for a given initial velocity and gravitational acceleration, the range depends on sin 2θ.

The sine function reaches its maximum value of $1$ when its argument is $90\degree$ .
For $sin2\theta$ to equal $1$ , $2\theta$ must be $90\degree$ .
Solving for $\theta$ : $\theta=45\degree$

Therefore, a 45° launch angle maximises the range of a projectile.

Practical Implications

Symmetry: At 45°, the time spent ascending equals the time descending, maximising horizontal travel.
Balance: This angle optimally balances the trade-off between vertical height and horizontal distance.

Practice Question 1

A baseball is hit with an initial velocity of 40 m/s. Calculate:

The maximum range
The range if hit at 30°

Maximum range (at 45°):

R = \frac{v^2 \sin 2\theta}{g} \\= \frac{(40)^2 \sin (2 \cdot 45°)}{9.8} \\= 163.3m

Range at 30°:

R = \frac{(40)^2 \sin (2 \cdot 30°)}{9.8} \\= 141.4m

This example demonstrates that the 45° angle indeed provides the maximum range.

What are the Factors Affecting Optimal Angle?

While 45° is theoretically optimal, real-world scenarios may differ due to:

Air resistance: Can reduce the optimal angle to around 40-43°.
Elevation differences: If the landing point is higher or lower than the launch point, the optimal angle changes.
Wind: Can affect both the optimal angle and the trajectory.

Maximum Height, Time of Flight, and Range

Up Next

Mastering Projectile Motion: Practice Problems and Solutions

Return to Module 5: Advanced Mechanics