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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)
Explanation
We know, $R = 4H \cot \theta \Rightarrow \cot \theta = 1/2$ From triangle we can say that $\sin \theta = 2/\sqrt{5}$, $\cos \theta = 1/\sqrt{5}$ $\therefore$ Range of projectile $R = 2v^{2}\sin \theta \cos \theta/g$ [add image] = $(2v^{2}/g) \times (2/\sqrt{5}) \times (1/\sqrt{5}) = 4v^{2}/5g$