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Question

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


Solution

 

max height ,H= v2sin2θ/2g

velocity at max height = ux= vcosθ

angular momentum= muxH

and putting θ =450

angular momentum = mv3/g4√2

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