Prove that the maximum horizontal range is four times the maximum height attained by the projectile, when fired at an inclination so as to have maximum horizontal range.

Explanation

For `theta = 45^(@)`, the horizontal range is maximum and is given by
`R_(max) = (u^(2))/(g)`
Maximum height attained,
`H_(max) = (u^(2) sin^(2) 45^(@))/(2g) = (u^(2))/(4g) = (R_(max))/(4)`
or `R_(max) = 4H_(max)`