Question

Show that the projection angle $0^o$ for a projectile launched origin is given by
${\theta }_0={\tan }^{-1}\left(\frac{4hm}{R}\right)$
where, the symbols have their usual meanings.

Answer 1

The maximum vertical height is given as,

$h_m=\frac{u_0{\sin }^2{\theta }_0}{2g}$ (1)

The horizontal range is given as,

$R=\frac{u_0{\sin }^{2}2{\theta }_0}{g}$ (2)

From equation (1) and (2), we get

$\frac{h_m}{R}=\frac{{\sin }^2{\theta }_0}{2{\sin }^{2}2{\theta }_0}$

$\frac{h_m}{R}=\frac{\sin {\theta }_0}{4\cos {\theta }_0}$

$\tan {\theta }_0=\frac{4h_m}{R}$

${\theta }_0={\tan }^{-1}\left(\frac{4h_m}{R}\right)$