Projection at Angle from Height — Range

The situation

A body launched at angle $\theta$ from height $h$ with speed $u$. It lands on the ground below. The horizontal distance from directly below the launch point to the landing point is the range $R$.

Derivation

The horizontal motion is simple — constant velocity, no acceleration:

$R = u_x \times T = u\cos\theta \times T$

$\boxed{R = u\cos\theta \cdot T}$

where $T$ is the time of flight:

$T = \frac{u\sin\theta + \sqrt{u^2\sin^2\theta + 2gh}}{g}$

Substituting:

$R = \frac{u\cos\theta}{g}\left(u\sin\theta + \sqrt{u^2\sin^2\theta + 2gh}\right)$

Why there is no single clean formula

For the standard projectile (from ground level), the time of flight simplified to $\frac{2u\sin\theta}{g}$, giving a clean range formula $R = \frac{u^2\sin2\theta}{g}$.

Here, $T$ involves a square root that cannot be simplified further. So $R = u\cos\theta \cdot T$ is the practical formula — compute $T$ first, then multiply by $u\cos\theta$.

The optimal angle is no longer 45°

For the standard projectile, maximum range is at $\theta = 45°$. When launching from a height, the optimal angle shifts below 45° — because a lower angle gives more horizontal velocity, and the extra height already provides sufficient time in the air.

The exact optimal angle depends on $h$ and $u$ and requires calculus to find. For large $h$ relative to $u^2/g$, the optimal angle approaches $0°$ (horizontal projection gives the most range).

Approach for problems

1. Identify $u$, $\theta$, $h$
2. Compute $T$ from the vertical equation: $-h = u\sin\theta \cdot T - \frac{1}{2}gT^2$, solve the quadratic, take the positive root
3. Compute $R = u\cos\theta \cdot T$