Q 11-03-011NEETJEE MainAIPMT 2010Top questionMedium
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 the acceleration due to gravity)
Answer: (A) $\dfrac{4v^2}{5g}$
Concept. $R = \dfrac{2v^2\sin\theta\cos\theta}{g}$, $H = \dfrac{v^2\sin^2\theta}{2g}$.
Step 1. Angle. $R = 2H \Rightarrow 2\sin\theta\cos\theta = \sin^2\theta \Rightarrow \tan\theta = 2$.
Step 2. Range. $\sin\theta = \dfrac{2}{\sqrt5}$, $\cos\theta = \dfrac{1}{\sqrt5}$:
$$R = \frac{2v^2}{g} \cdot \frac{2}{5} = \frac{4v^2}{5g}$$
Answer. $\dfrac{4v^2}{5g}$ (Option A).
Solution by Sreeraj P, M.Sc Physics