The horizontal range of a projectile is `2sqrt(3)` time its maximum height. Find the angle of projection .

Explanation

If u and `alpha` be the initial velocity of projection and angle of projection respectively, then the maximum height attaiend ,`H_(m) = (u^(2) sin^(2) alpha)/(2g)` and horizontal range , `R = (2u^(2) sin alpha cos alpha)/(g)`
According to the problem , we can write
`(2u^(2) alpha cos alpha)/(g) = 2sqrt(2) ((u^(2) sin^(2) alpha)/(2g)) rArr tan alpha = ((2)/(sqrt(3))) rArr alpha = tan^(-1)((2)/(sqrt(3)))`