If R is the horizontal range for `theta` inclination and h is the maximum height reached by the projectile, show that the maximum ragne is given by `(R^(2))/(8h) + 2h`.

Explanation

We know that horizontal range .
`R = (u^(@) sin 2theta)/(g)` and maximum height , `h = (u^(2) sin^(2) theta)/(2g)`
`therefore (R^(2))/(8h) +2h = ([(u^(2)sin 2 theta)/(g)]^(2))/(8[(u^(2) sin^(2) theta)/(2g)]) + 2 [(u^(2) sin^(2)theta)/(2g)]`
` = (u^(4) (2 sin theta cos theta)^(2))/(g^(2) xx 8 (u^(2) sin theta)/(2g)) + (u^(2) sin^(2) theta)/(g)`
` = (u^(2))/(g) (cos^(2) theta + sin^(2) theta) = (u^(2))/(g)= R_(max)`