An object of mass m is projected with a momentum ρ at such an angle that its maximum height is 1/4th ofits horizontal range. Its minimum kinetic energy in its path will be

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p28m

p24m

3p24m

p2m

answer is B.

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Detailed Solution

It is given that, H=R4∴ u2sin2θ2g=2u2sinθcosθ4g or tanθ=1⇒θ=450At highest point, momentum will remain p2∴ K=(p/2)22m=p24m

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