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Question

A particle of mass m is projected with a velocity v making an angle of 45 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 given by which of the following relation(s)?
  1. zero
  2. mv342 g
  3. mv32 g
  4. m2gh3


Solution

The correct options are
B mv342 g
D m2gh3
Angular momentum(L) = r×p
L = mvrsinθ
L= (momentum)×(perpendicular distance)


Angular momentum about origin

L=mv2 h    ----(i)
h= max. height of a projectile
h=v2 sin2θ2 g
h=v24 g   { θ=45}   ----(ii)

from (i) & (ii),

L=m2(2gh h) = m2gh3

L=mv2×v24g=mv342 g 

Option B and D are correct.

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