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Q.

A particle of mass m is projected with a velocity v making an angle of 45o with the horizontal. The magnitude of angular momentum of the projectile about an axis of projection when the particle is at maximum height h is

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a

0

b

mv3/42g

c

mv2/2g

d

m2gh3

answer is B.

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Detailed Solution

0−v2sin2⁡45∘=2gh or h=v2/4gAt highest point, momentum=mvcos⁡45∘=mv/2Angular momentum=mv2×h=mv2×v24g=mv342gIn terms of h, angular momentum=mv2×h=m2[(4gh)]×h=m2gh3
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