Let $\phi $ be elevation angle of the projectile at its highest point as seen from the point of projection O and $\theta $ be angle of projection with the horizontal.

From figure, tan$\phi $ = ${H \over {R/2}}$      ....(1)

In case of projectile motion

Maximum height H = $${{{u^2}{{\sin }^2}\theta } \over {2g}}$$

Horizontal range, R = $${{{u^2}\sin \theta } \over g}$$

Substituting these values of H and R in (1), we get

$$\tan \phi = {{{{{u^2}{{\sin }^2}\theta } \over {2g}}} \over {{{{u^2}\sin 2\theta } \over {2g}}}}$$

$$\tan \phi = {{{{\sin }^2}\theta } \over {\sin 2\theta }} = {{{{\sin }^2}\theta } \over {2\sin \theta \cos \theta }}$$

= $${1 \over 2}\tan \theta $$

$$\tan \phi = {1 \over 2}\tan 45^\circ = {1 \over 2}$$

Here, $\theta $ = 45°

$ \therefore $ $$\tan \phi = {1 \over 2}\tan 45^\circ = {1 \over 2}$$    $$ \left(\because {\tan 45^\circ = 1} \right)$$

$$\phi = {\tan ^{ - 1}}\left( {{1 \over 2}} \right)$$