Question

Find the angle of projection for which the maximum height of the projectile is double the horizontal range attained by the body.

Answer 1

Let $\theta$ be the angle of projection for which the horizontal range of the projectile are double

We know that

Formula of height

$H=\frac{u^2{\sin }^2\theta }{2g}$

Formula of range

$R=\frac{u^2\sin 2\theta }{g}$

Now, if maximum height is double of range

$H=2R$

$\frac{u^2{\sin }^2\theta }{2g}=2\times \frac{u^2\sin 2\theta }{g}$

${\sin }^2\theta =4\left(2\sin \theta \cos \theta \right)$

$\tan \theta =8$

$\theta ={\tan }^{-1}\left(8\right)$

Hence, the angle of projection is ${\tan }^{-1}\left(8\right)$