Question

A particle is projected with velocity $v$ such that its range on the horizontal plane is twice the greatest height attained by it,The range attained by it, The range of the projectile is (when it is acceleration due to gravity is ${}^{\prime }{g}^{\prime }$)

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

Answer 1

The correct option is A $\frac{4v^2}{5g}$

$2\times \frac{4^2\sin \theta \cos \theta }{g}=2\times \frac{4^2{\sin }^2\theta }{2g}$

$2\cos \theta =\sin \theta$

$\Rightarrow \tan \theta =2$

$\Rightarrow \sin \theta =\frac{2}{\sqrt{5}};\cos \theta =\frac{1}{\sqrt{5}}$

$\therefore R=2\times \frac{4^2}{g}\frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}$

$=\frac{4v^2}{5g}$

Hence,

option $\left(A\right)$ is correct answer.