A particle is projected with a velocity v so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is

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4v25g

4 g5 V2

4v35g2

4v5 g2

answer is A.

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Detailed Solution

Since R = 1Hv2sin2θg=2×v2sin2θ2 g 2sinθcosθ=sin2θ or tan θ=2 R=v2sin2θg =v22sinθcosθg=2v2 g×25×15=4v25 g

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