The angle of projection, of which the horizontal range and maximum height of projectile are equal is

We know that

$\mathrm{R}=\frac{{\mathrm{u}}^2\sin 2\mathrm{\theta }}{\mathrm{g}}\text{ and }{\mathrm{h}}_{max}=\frac{{\mathrm{u}}^2{\sin }^2\mathrm{\theta }}{2\mathrm{g}}$

Given that $\mathrm{R}={\mathrm{h}}_{max}$

$\begin{array}{l}\therefore \frac{{\mathrm{u}}^2\sin 2\mathrm{\theta }}{\mathrm{g}}=\frac{{\mathrm{u}}^2{\sin }^2\mathrm{\theta }}{2\mathrm{g}}\\ \text{ or }2\sin \mathrm{\theta cos}\mathrm{\theta }=\left({\sin }^2\mathrm{\theta }/2\right)\\ \text{ or }\tan \mathrm{\theta }=4\text{ or }\mathrm{\theta }={\tan }^{-1}\left(4\right)\end{array}$