⚽ Launch Angle Analysis

The range of a projectile is the total horizontal displacement from launch to landing. For a projectile launched on level ground with speed $v$ at angle $\theta$ above horizontal, we can derive a compact formula for the range.

Launch Angle Analysis

Derivation

The horizontal velocity is constant (no air resistance):

$v_x = v\cos\theta$

The time of flight is found from vertical motion. On level ground, the projectile returns to its initial vertical position, so $s_y = 0$:

$s_y = v_y t + \tfrac{1}{2}a_y t^2$

$0 = v\sin\theta \cdot t_{\text{flight}} - \tfrac{1}{2}g \cdot t_{\text{flight}}^2$

$t_{\text{flight}} = \dfrac{2v\sin\theta}{g}$

The range is horizontal velocity multiplied by time of flight:

$R = v_x \times t_{\text{flight}} = v\cos\theta \times \dfrac{2v\sin\theta}{g} = \dfrac{v^2(2\sin\theta\cos\theta)}{g}$

Using the double-angle identity $\sin(2\theta) = 2\sin\theta\cos\theta$:

Maximum Range

The maximum value of $\sin(2\theta)$ is 1, which occurs when $2\theta = 90°$, giving $\theta = 45°$.

At $\theta = 45°$:

$R_{\text{max}} = \dfrac{v^2}{g}$

This is a remarkable result: the maximum range depends only on launch speed and gravitational acceleration, not on any other parameter.

A powerful symmetry in the range equation reveals that two different launch angles can produce exactly the same range.

Recall the identity: $\sin(180° - \phi) = \sin(\phi)$. If we set $\phi = 2\theta$, then:

$\sin(2\theta) = \sin(180° - 2\theta) = \sin\bigl(2(90° - \theta)\bigr)$

This means launching at angle $\theta$ gives the same range as launching at angle $(90° - \theta)$. These angles are called complementary angles (they sum to 90 degrees).

So far we have assumed launch and landing at the same height. In many real situations this is not true — a projectile may be launched from a cliff, a ramp, or a building, or may land on a slope.

Time of Flight from Height

When a projectile is launched from height $h$ above the landing point, the vertical displacement is $s_y = -h$ (downward). Using the quadratic formula on $s_y = v_y t + \tfrac{1}{2}a_y t^2$:

$-h = v\sin\theta \cdot t - \tfrac{1}{2}g \cdot t^2$

$\tfrac{1}{2}gt^2 - v\sin\theta \cdot t - h = 0$

Solving this quadratic for the positive root:

$t_{\text{flight}} = \dfrac{v\sin\theta + \sqrt{v^2\sin^2\theta + 2gh}}{g}$

The extra term $+2gh$ under the square root increases the flight time compared to level ground. The range is then $R = v\cos\theta \times t_{\text{flight}}$.

Why the Optimal Angle Changes

Complete the table for a projectile launched at $v = 20\ \text{m/s}$ on level ground. Take $g = 9.8\ \text{m/s}^2$.

Questions:

1. Which angle gives the maximum range? By how much does it exceed the range at 15 degrees?
2. Which pairs of angles give identical ranges? Explain why using the complementary angle property.
3. As the angle increases from 15 degrees to 45 degrees, does the time of flight increase or decrease? Explain.

A projectile is launched from the top of a 50 m cliff at $18\ \text{m/s}$. Calculate the range at each of the following angles and determine which gives the maximum range. Take $g = 9.8\ \text{m/s}^2$.

Use the formula:

$t_{\text{flight}} = \dfrac{v\sin\theta + \sqrt{v^2\sin^2\theta + 2gh}}{g}$

then $R = v\cos\theta \times t_{\text{flight}}$.

Conclusion: What is the optimal launch angle from this cliff? Is it greater than, less than, or equal to 45 degrees? Explain why in terms of horizontal velocity and flight time.