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Question

A particle of mass m is projected with velocity v at an angle θ with the horizontal. Find its angular momentum about the point of projection when it is at the highest point of its trajectory.

A
mv2sin2θcosθ2g
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B
mv3sinθcosθg
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C
mv3sin2θcosθ2g
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D
mv3sin2θcosθg
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Solution

The correct option is C mv3sin2θcosθ2g
At the highest point, it has horizontal velocity only i.e. vx=vcosθ.
Length of the perpendicular to the horizontal velocity from O is the maximum height, where
Hmax=v2sin2θ2g

We know,
L=mrv
=m×Hmax×vcosθ
=m×(v2sin2θ2g)×vcosθ
=mv3sin2θcosθ2g
Option (c) is correct.

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