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A body of mass m is projected with a velocity u at an angle of 60º with the horizontal. The magnitude of the angular momentum about the point of projection when the body is at the highest point of its trajectory is

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mu32g

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3mu316g

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detailed solution

Correct option is C

At highest pointv=ucos⁡60∘=u2Hence at maximum height perpendicular distance from line of projection =u2sin2⁡θ2g=3u28gL=mvr=mucos⁡60∘×3u28g=3mu316g

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