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)
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R = 2H given
We know \(R = 4H \cot \theta\) ⇒ ⇒ \(\cot \theta = \frac{1}{2}\)
From triangle we can say that \(\sin \theta = \frac{1}{\sqrt{5}}\) , \(\cos \theta = \frac{1}{\sqrt{5}}\)
∴ ∴ Range of projectile \(R = \frac{v^{2} \sin \theta \cos \theta}{g}\)
\(= \frac{2v^{2}}{g} \times \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}}\) = \(\frac{4v^{2}}{5g}\) .
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