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

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a

zero

b

mu3/42g

c

mu3/2g

d

m2gh3

answer is B.

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Detailed Solution

ux=ucos⁡45∘=u/2Linear momentum =mu/2Maximum height =u2/4gAngular momentum of the particle= moment of momentum =mu2×u24g=mu342g

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