Question

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.

Answer 1

$R=\frac{u^{2}sin2\theta }{g},H=\frac{u^2{sin}^2\theta }{2g}{R}_{max}=\frac{u^2}{g}for\theta ={45}^0,thus,forsame\theta ,H=\frac{u^2{sin}^2{45}^0}{2g}=\frac{u^2}{4g}\frac{R_{max}}{H}=4$