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.
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:
$-\frac{h_{ m }}{R}=\frac{\sin ^{2} \theta}{2 \sin ^{2} \theta}$
$=\frac{\sin \theta \times \sin \theta}{2 \times 2 \sin \theta \cos \theta}$
$=\frac{1}{4} \frac{\sin \theta}{\cos \theta}=\frac{1}{4} \tan \theta$
$\tan \theta=\left(\frac{4 h_{ m }}{R}\right)$
$\theta=\tan ^{-1}\left(\frac{4 h_{ m }}{R}\right)$
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