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Question

# A particle of mass 'm' is projected with velocity 'v' 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

along positive z Axis

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B
along negative z Axis
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C

along positive z Axis

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D

along negative z Axis

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Solution

## The correct option is D along negative z Axis At the highest point, it has only horizontal velocity VX=vcosθ Now length of the perpendicular to the horizontal velocity from 'O' is the maximum height reached. We already know, Hmax = ∴ The required angular momentum L=mv3sin2θcosθ2g, which is directed along negative z axis

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