# Find the angle of projection at which the horizontal range and the maximum height of a projectile are equal.

Let the required angle of projection be $\theta$ for a projectile fired with velocity $u$. According to the question,

$\text{Maximum Height}=\text{Horizontal Range}$ $\frac{u^2\sin^2\theta}{2g}=\frac{u^2\sin2\theta}{g}$ $\frac{\sin^2\theta}{2}=\sin2\theta$ $\frac{\sin^2\theta}{2}=2\sin\theta\cos\theta$ $\frac{\sin\theta}{\cos\theta}=4$ $\tan\theta=4$ $\therefore\theta=\tan^{-1}4=75.96°$

Hence, the required angle of projection is $75.96°.$