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A particle is projected with a velocity $v$ , so that its range on a horizontal plane is twice the greatest height attained. Then its range is

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4v25g

4g5v2

4v5g

2v25g

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detailed solution

Correct option is A

R=2H ⇒ 2v2sinθcosθg=2(V2sin2θ2g) ⇒ tanθ=2 ⇒ R=2gV2(25)(15)=4V25g

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A particle is projected with a velocityv , so that its range on a horizontal plane is twice the greatest height attained. Then its range is

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A particle is projected with a velocity $v$ , so that its range on a horizontal plane is twice the greatest height attained. Then its range is