Question

Prove that maximum range of a projectile is 4 times its maximum height.

Answer 1

\(\because \ =R_{max} = \frac{u^2}{g}\)

It is obtained at θ = 45°, therefore at angle of projection

\(H_{max} = \frac{u^2\sin^2 45^\circ}{2g} = \frac{u^2\times\frac{1}{2}}{2g} = \frac{u^2}{4g}\)

or  \(H_{max} = \frac{1}{4}\times\frac{u^2}{g} = \frac{1}{4}\times R_{max}\)

\(\therefore\)  \(R_{max} = 4H_{max}\)