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 $75.96°.$
[Read: Projectile Motion]
