Launch Angle Analysis

Deriving a compact formula for horizontal range on level ground

Fire a cannonball from level ground at 100 m/s. At 30° it lands closer than at 45°; at 60° it also lands closer than at 45° but travels the same distance as the 30° shot. There is a single angle that maximises horizontal range when launch and landing height are equal, and deriving why requires combining the horizontal and vertical motion equations.

Launch Angle Analysis, how angle affects range on level ground.

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 $\theta = 45°$:

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

Within these model conditions, the maximum range depends only on launch speed and gravitational acceleration.

For a point projectile with equal launch and landing heights, uniform $g$ and negligible drag, $R = v^2\sin(2\theta)/g$; it is maximised at $\theta = 45°$, giving $R_{\text{max}} = v^2/g$. The time of flight is $t_{\text{flight}} = 2v\sin\theta/g$ and is derived by setting $s_y = 0$.

Pause, copy the highlighted range equation and the maximum-range result into your book before moving on.

Two different launch angles that produce exactly the same range

We just saw that the range equation contains $\sin(2\theta)$, which peaks at 45°. That raises a question: are there other angles that give the same range as each other? This card answers it → complementary angles (summing to 90°) produce identical ranges.

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°).

Complementary angles $\theta$ and $(90° - \theta)$ produce identical ranges on level ground because $\sin(2\theta) = \sin(180° - 2\theta)$; for example, 30° and 60° always give the same range at any launch speed.

Add the highlighted complementary-angle rule to your notes before the check below.

How height difference shifts the optimal launch angle away from 45°

We just saw that 45° maximises range on level ground and complementary angles give equal ranges. That raises a question: does 45° stay optimal when launch and landing heights differ? This card answers it → height difference shifts the optimal angle below 45° (from a cliff) or above 45° (uphill).

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

When launched from height $h$, flight time is $t = (v\sin\theta + \sqrt{v^2\sin^2\theta + 2gh})/g$; launching above landing makes the optimal angle less than 45°, and launching below landing (uphill) makes it greater than 45°.

Pause, write the highlighted cliff-launch formula and the two height-effect rules into your book.

Component independence and the height effect, measured by experiment instead of asserted

Misconceptions, final check