Q.

A particle of mass m is projected with a velocity v making an angle of 450 with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at maximum height h is mv3(xg). Find the value of x.

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answer is 0032.00.

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Detailed Solution

At the highest point, velocity =vcos450=v2 in horizontal direction ​∴ Momentum =mv2​L= Angular momentum = Momentum × Perpendicular distance ​⇒L=mv2×h​ Here h=v2sin24502g=v24g​∴L=mv2v24 g=mv342 g=mv332 g
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A particle of mass m is projected with a velocity v making an angle of 450 with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at maximum height h is mv3(xg). Find the value of x.