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.

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Solution

max height ,H= v²sin²θ/2g

velocity at max height = u_x= vcosθ

angular momentum= mu_xH

and putting θ =45⁰

angular momentum = mv³/g4√2

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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.

max height ,H= v2sin2θ/2g velocity at max height = ux= vcosθ angular momentum= muxH and putting θ =450 angular momentum = mv3/g4√2

max height ,H= v²sin²θ/2g

velocity at max height = u_x= vcosθ

angular momentum= mu_xH

and putting θ =45⁰

angular momentum = mv³/g4√2