A particle is projected with a velocity u. So that its horizontal range and maximum height reached are equal. The maximum height reached is

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2u2/3g

4u2/5g

u2/g

8u2/17g

answer is D.

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Detailed Solution

R=Hmax2u2sinθcosθg=u2sin2θ2gtanθ=4sinθ=417Hmax=u2.16/172g=8u217g

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