Show that for a projectile the angle between the velocity and the $x$ -axis as a function of time is given by

$\theta(t)=\tan ^{-1}\left(\frac{v_{0 y}-g t}{v_{0 x}}\right)$

Show that the projection angle $\theta_{0}$ for a projectile launched from the origin is given by

$\theta_{0}=\tan ^{-1}\left(\frac{4 h_{m}}{R}\right)$

Where the symbols have their usual meaning.

Let $v_{0 x }$ and $v_{0y }$ respectively be the initial components of the velocity of the projectile along horizontal $(x)$ and vertical $(y)$ directions.

Let $v_{x}$ and $v_{y}$ respectively be the horizontal and vertical components of velocity at a point $P$

Time taken by the projectile to reach point $P =t$ Applying the first equation of motion along the vertical and horizontal directions, we get $v_{y}=v_{0 y}= g t$

And $v_{x}=v_{0 x}$

$\therefore \tan \theta=\frac{v_{y}}{v_{x}}=\frac{v_{0 y}- g t}{v_{0 x}}$

$\theta=\tan ^{-1}\left(\frac{v_{0 y}-g t}{v_{0 x}}\right)$

Maximum vertical height, $h_{ m }=\frac{u_{0}^{2} \sin ^{2} 2 \theta}{2 g }$

Horizontal range, $\quad R=\frac{u_{0}^{2} \sin ^{2} 2 \theta}{g}$

Solving equations $(i)$ and $(ii)$, we get:

$-\frac{h_{ m }}{R}=\frac{\sin ^{2} \theta}{2 \sin ^{2} \theta}$

$=\frac{\sin \theta \times \sin \theta}{2 \times 2 \sin \theta \cos \theta}$

$=\frac{1}{4} \frac{\sin \theta}{\cos \theta}=\frac{1}{4} \tan \theta$

$\tan \theta=\left(\frac{4 h_{ m }}{R}\right)$

$\theta=\tan ^{-1}\left(\frac{4 h_{ m }}{R}\right)$