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 (when it is acceleration due to gravity is ‘g’) [PMT 2010]

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$\frac{4 v^{2}}{5 g}$

$\frac{v^{2}}{g}$

$\frac{4 g}{5 v^{2}}$

$\frac{4 v^{2}}{\sqrt{5 g}}$

answer is A.

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Detailed Solution

$Given R = 2 H; T h e r e f o r e Tan θ = 2 a n d \sin 2 θ = \frac{2 \tan θ}{1 + \tan^{2} θ} = \frac{4}{5} . T h e r e f o r e R = \frac{u^{2} Sin 2 θ}{g}$ .

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