Answer to Question #96926 in Mechanics | Relativity for Dude

Question

Need an explanation please.

Prove it mathematically and in words why a launch angle of 45 degrees will provide the max range of a projectile.

Answer 1

The horizontal range of a projectile is $d = \frac{v^2 sin 2\theta}{g}$.

For a fixed $v$ and $g$, in order to maximize $d$, we need to maximize $\sin 2 \theta$. Since $\sin x$ attains its maximum value at $x = \frac{\pi}{2}$, $\sin 2 \theta$ is maximized at $\theta = \frac{\pi}{4}$.