Question

A particle is projected with a velocity $v$ so that its horizontal range twice the greater height attained. The horizontal range is

• A. $\frac{v^2}{g}$
• B. $\frac{2v^2}{3g}$
• C. $\frac{4v^2}{5g}$
• D. $\frac{v^2}{2g}$

Answer 1

The correct option is A $\frac{v^2}{g}$

Horizontal range $R=\frac{v^2\sin 2\theta }{g}$

Maximum height $H=\frac{v^2{\sin }^2\theta }{2g}$

But $R=2H$

$\therefore$ $\frac{v^2\sin 2\theta }{g}=$ $2\frac{v^2{\sin }^2\theta }{2g}$

We get, $\sin 2\theta =2{\sin }^2\theta$

Or $2\sin \theta \cos \theta =2{\sin }^2\theta$

Or $\tan \theta =1$ $⟹\theta ={45}^o$

Thus horizontal range $R=\frac{v^2\sin \left({90}^o\right)}{g}=\frac{v^2}{g}$