A projectile is fired at an angle of ${45}^0$ with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is

Height of projectile

$\mathrm{H} = \frac{{\mathrm{u}}^2{\sin }^2\mathrm{\theta }}{2\mathrm{g}} = \frac{{\mathrm{u}}^2{\sin }^2{45}^0}{2\mathrm{g}}$

$\Rightarrow \mathrm{H} = \frac{{\mathrm{u}}^2}{4\mathrm{g}} ------\left(\mathrm{i}\right)$

Range of projectile R = $\frac{{\mathrm{u}}^2\sin 2\mathrm{\theta }}{\mathrm{g}} = \frac{{\mathrm{u}}^2\sin {90}^0}{\mathrm{g}}$

$\Rightarrow \mathrm{R} = \frac{{\mathrm{u}}^2}{\mathrm{g}}$

      $\tan \mathrm{\alpha } = \frac{\mathrm{H}}{{\displaystyle \frac{\mathrm{R}}{2}}}$

In $∆\mathrm{OAB} = \frac{{\displaystyle \frac{{\mathrm{u}}^2}{4\mathrm{g}}}}{{\displaystyle \frac{{\mathrm{u}}^2}{2\mathrm{g}}}}$

    tan $\mathrm{\alpha } =\frac{1}{2} \Rightarrow \mathrm{\alpha } = {\tan }^{-1}\left(\frac{1}{2}\right)$