(a) Let $v_{ox}$ and $v_{oy}$ be the initial component velocity of the projectile at O along OX direction and OY direction respectively, where OX is horizontal and the OY is vertical. Let the projectilego from O to P in time t and $v_x, v_y$ be the component velocity of projectile at P along horizontal and vertical directions respectively. Then, $v_y = v_{oy} - gt$ and $v_x = v_{ox}$If θ is the angle which the resultant velocity $\vec{v}$ makeswith horizontal direction, then$\tan \theta = \frac{v_y}{v_x} = \frac{v_{oy} - gt}{v_{ox}}$ or $\theta = \tan^{-1}\left(\frac{v_{oy} - gt}{v_{ox}}\right)$(b) In angular projection, Maximum vertical height, $h_m = \frac{u^2 \sin^2 \theta_0}{2g}$Horizontal range,$R = \frac{u^2 \sin 2\theta_0}{g} = \frac{u^2}{g} 2 \sin \theta_0 \cos \theta_0$So, $\frac{h_m}{R} = \frac{\tan \theta_0}{4}$ or $\tan \theta_0 = \frac{4 h_m}{R}$ or $\theta_0 = \tan^{-1}\left(\frac{4 h_m}{R}\right)$