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)?zeromv34√2 gmv3√2 gm√2gh3

Solution

## The correct options are B mv34√2 g D m√2gh3Angular momentum(L) = →r×→p L = mvrsinθ L= (momentum)×(perpendicular distance) Angular momentum about origin L=mv√2 h    ----(i) h= max. height of a projectile h=v2 sin2θ2 g h=v24 g   {∵ θ=45∘}   ----(ii) from (i) & (ii), L=m√2(2√gh h) = m√2gh3 L=mv√2×v24g=mv34√2 g  Option B and D are correct.

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