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

In angualr projection fo projectile , the horizontal range is mabimum, when `theta =45^(@).
`R_(max) =(u^(2) sin 2 xx 445^(@))/g =u^(2)/g`
Asx. Height,
`Hmax) =(u^(2) sin ^(2) 45^(@))/ (2 g) =(u^(2) (1// sqrt2) ^(2))/(2 g)`
`= u^(2)/(4 g) =R_max)/4`
:. ` R_(max) =4 H_(max) (Prpved)`