A body is projected with velocity u such that its horizontal range and maximum vertical heights are same. The maximum heights is

A body is projected with velocity = u

given that horizontal range $\text{R}=\frac{{\text{u}}^{\text{2}}\sin 2\theta }{\text{g}}$

and maximum vertical height $=\frac{{\text{u}}^{\text{2}}{\sin }^2\theta }{\text{2g}}$

are same then$\text{R}={\text{h}}_{\max }$

$\frac{{\text{u}}^{\text{2}}\sin 2\theta }{\text{g}}=\frac{{\text{u}}^{\text{2}}{\sin }^2\theta }{\text{2g}}$

$\Rightarrow 2\sin \theta ={\sin }^2\theta$

$\tan \theta =2$

then$\sin \theta =\frac{2}{\sqrt{5}}$

hence ${\text{H}}_{\max }=\frac{{\text{u}}^{\text{2}}{\sin }^2\theta }{\text{2g}}=\frac{{\text{u}}^{\text{2}}}{\text{2g}}\frac{4}{5}=\frac{{\text{2u}}^{\text{2}}}{\text{5g}}$

(or)

$\text{H}=\text{R}$

$\text{Tan}\theta =\frac{\text{4H}}{\text{R}}=\frac{4}{1}$

$\sin \theta =\frac{4}{\sqrt{17}}$

$\text{H}=\frac{{\text{u}}^{\text{2}}{\sin }^2\theta }{2\text{g}}=\frac{{\text{u}}^2\frac{16}{17}}{2\text{g}}=\frac{{\text{8u}}^{\text{2}}}{\text{17g}}$