# For what angle range becomes double of height in a projectile motion?

Jun 9, 2018

${\tan}^{-} 1 2$

#### Explanation:

The range for projectile is $R = {u}^{2} / g \sin \left(2 \alpha\right)$
and the maximum height is given by $H = \frac{{u}^{2} {\sin}^{2} \alpha}{2 g}$

Thus, to get $R = 2 H$ we must have

${u}^{2} / g \sin \left(2 \alpha\right) = 2 \frac{{u}^{2} {\sin}^{2} \alpha}{2 g} \implies$

$\sin \left(2 \alpha\right) = {\sin}^{2} \alpha \implies$

$2 \sin \alpha \cos \alpha = {\sin}^{2} \alpha \implies$

$\tan \alpha = 2$

Thus, the angle of projection must be ${\tan}^{-} 1 2$