Question

For a projectile if the maximum height attained is equal to horizontal range, then it's angle of projection with vertical is

Answer 1

Answer (4)
$\frac{2{\sin }^2\theta }{g}=\frac{u^{2}2{\sin }^2\theta }{2g}$
(Where $\theta$ is from horizontal)
$\tan \theta =4$
$\theta ={\tan }^{-1}\left(4\right)$
Angle with vertical $=\frac{\pi }{2}-\theta$