Show that for a projectile,the angle between the velocity and the $x$-axis as a function of time is given by $\theta(t) = \tan^{-1}\left(\frac{v_{0y} - gt}{v_{0x}}\right)$.
Show that the projection angle $\theta_0$ for a projectile launched from the origin is given by $\theta_0 = \tan^{-1}\left(\frac{4h_m}{R}\right)$,where the symbols have their usual meanings.

(N/A) Let $v_{0x}$ and $v_{0y}$ be the initial components of the velocity of the projectile along the horizontal $(x)$ and vertical $(y)$ directions,respectively.
Let $v_x$ and $v_y$ be the horizontal and vertical components of velocity at any time $t$.
Using the first equation of motion:
$v_x = v_{0x}$
$v_y = v_{0y} - gt$
The angle $\theta$ with the $x$-axis is given by:
$\tan \theta = \frac{v_y}{v_x} = \frac{v_{0y} - gt}{v_{0x}}$
$\theta(t) = \tan^{-1}\left(\frac{v_{0y} - gt}{v_{0x}}\right)$
For a projectile launched with initial velocity $u_0$ at angle $\theta_0$:
Maximum height $h_m = \frac{u_0^2 \sin^2 \theta_0}{2g} \quad (i)$
Horizontal range $R = \frac{u_0^2 \sin(2\theta_0)}{g} = \frac{2u_0^2 \sin \theta_0 \cos \theta_0}{g} \quad (ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{h_m}{R} = \frac{u_0^2 \sin^2 \theta_0}{2g} \times \frac{g}{2u_0^2 \sin \theta_0 \cos \theta_0}$
$\frac{h_m}{R} = \frac{\sin \theta_0}{4 \cos \theta_0} = \frac{1}{4} \tan \theta_0$
Therefore,$\tan \theta_0 = \frac{4h_m}{R}$
$\theta_0 = \tan^{-1}\left(\frac{4h_m}{R}\right)$