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) $\dfrac{4v^2}{5g}$
$$R = 2H \;\Rightarrow\; \frac{2v^2\sin\theta\cos\theta}{g} = 2\cdot\frac{v^2\sin^2\theta}{2g} \;\Rightarrow\; \tan\theta = 2$$
So $\sin\theta = \dfrac{2}{\sqrt{5}}$ and $\cos\theta = \dfrac{1}{\sqrt{5}}$:
$$R = \frac{2v^2}{g}\cdot\frac{2}{\sqrt{5}}\cdot\frac{1}{\sqrt{5}} = \frac{4v^2}{5g}$$
Solution by Sreeraj P, M.Sc Physics