Horizontal Projection — Range

The situation

A body is projected horizontally with speed $u$ from height $h$. It travels forward while falling. The horizontal distance from the launch point to where it lands is the range $R$.

Derivation

Range = horizontal velocity × time of flight.

Horizontal velocity is constant at $u$ (no horizontal acceleration).

Time of flight is $T = \sqrt{\frac{2h}{g}}$ (derived from vertical free fall through height $h$).

$R = u \times T = u \times \sqrt{\frac{2h}{g}}$

$\boxed{R = u\sqrt{\frac{2h}{g}}}$

What affects the range

$R = u\sqrt{\frac{2h}{g}}$

• Larger $u$ → larger $R$ (proportionally — double the speed, double the range)
• Larger $h$ → larger $R$ (more time in the air)
• Larger $g$ → smaller $R$ (falls faster, less time in the air)

Unlike the standard projectile, there is no angle to optimise — the launch is always horizontal. Range simply increases with speed and height.

Finding $u$ or $h$ from $R$

If range and height are given, find launch speed:

$u = \frac{R}{\sqrt{\frac{2h}{g}}} = R\sqrt{\frac{g}{2h}}$

If range and speed are given, find height:

$\sqrt{\frac{2h}{g}} = \frac{R}{u} \implies \frac{2h}{g} = \frac{R^2}{u^2} \implies h = \frac{gR^2}{2u^2}$

Velocity at landing

At landing, the body has:

• Horizontal: $v_x = u$
• Vertical: $v_y = gT = g\sqrt{\frac{2h}{g}} = \sqrt{2gh}$

Speed at landing:

$|\vec{v}| = \sqrt{u^2 + 2gh}$

This can also be obtained from energy conservation — a useful check.