If R and H represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal is

Maximum height, H = $\frac{{\mathrm{u}}^2{\sin }^2\mathrm{\theta }}{2\mathrm{g}} ------\left(\mathrm{i}\right)$

Horizontal range, R = $\frac{{\mathrm{u}}^2\sin 2 \mathrm{\theta }}{\mathrm{g}} = \frac{2{\mathrm{u}}^2 \sin \mathrm{\theta } \cos \mathrm{\theta }}{\mathrm{g}}$-------------(ii)

Dividing (i) by (ii), we get

$\frac{\mathrm{H}}{\mathrm{R}} = \frac{\tan \mathrm{\theta }}{4} \mathrm{or} \tan \mathrm{\theta } = \frac{4\mathrm{H}}{\mathrm{R}} \mathrm{or} \mathrm{\theta } = {\tan }^{-1}\left(\frac{4\mathrm{H}}{\mathrm{R}}\right)$