Find the angle of projection of a projectile for which the horizontal range and maximum height are equal.

Explanation

Let u be the initial velocity of a projectile, projected with an angle `theta` with the horizontal .
Given , Horizontal range =Maximum height
`(u^(2) sin 2 theta)/(g)=(u^(2) sin ^(2) theta)/(2g)`
`rArr 2 sin theta cos theta= (sin ^(2) theta)/(2)`
`rArr (sin theta)/(cos theta)=4` or `tan theta rArr theta=tan^(-1) (4)`
`rArr theta =75.96^(@)`