Q. A particle is projected with a velocity $v$ such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where $g$ is acceleration due to gravity)
A) $\frac{4v^{2}}{5g}$
B) $\frac{4g}{5v^{2}}$
C) $\frac{v^{2}}{g}$
D) $\frac{4v^{2}}{\sqrt{5}g}$
Explanation
(a) $R = 2H$ Given
We know $R = 4H\cot\theta \Rightarrow \cot\theta = \frac{1}{2}$
From triangle we can say that $\sin\theta = \frac{2}{\sqrt{5}},\cos\theta = \frac{1}{\sqrt{5}}$
$\therefore$ Range of projectile $R = \frac{2v^{2}\sin\theta\cos\theta}{g}$
$= \frac{2v^{2}}{g} \times \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}} = \frac{4v^{2}}{5g}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.