Question

If, for a given velocity, the maximum range is at a projection angle of 45, then there must be equal ranges for angles above and below this. Show this explicitly.

Answer 1

Explanation:

The range R of a projectile is given by

$R = \frac{v_0^2}{g} \sin 2\theta$

The maximum range $R_{max}$ occurs when $\sin 2\theta = 1\:\text{or}\:\theta = 45°$. Let $\alpha$ be the angle above or below 45°. Now let's look at the ranges brought about by these angle differences.

Case 1: Angle above 45°

We can write the range as

$R_+ = \dfrac{v_0^2}{g} \sin 2(45° + \alpha)= \dfrac{v_0^2}{g} \sin (90° + 2\alpha)$

$\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} (\sin 90° \cos 2\alpha + \cos 90° \sin 2\alpha)$

$\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} \cos 2\alpha\:\:\:\:\:(1)$

Case 2: Angle below 45°

We can write the range as

$R_- = \dfrac{v_0^2}{g} \sin 2(45° - \alpha)= \dfrac{v_0^2}{g} \sin (90° - 2\alpha)$

$\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} (\sin 90° \cos 2\alpha - \cos 90° \sin 2\alpha)$

$\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} \cos 2\alpha\:\:\:\:\:(2)$

Note that the equations (1) and (2) are identical. Therefore, the ranges are equal if they differ from 45° by the same amount.