Question

A particle of mass m is projected with velocity v making an angle of 45degree with the horizontal . The magnitude of the angular momentum of the Particle about the point of projection when the particle is at its maximum height is (g=acceleration due to gravity)

A) zero
B) mv^3/4√2g
C) mv^3/√2g
D) mv^2/2g

Answer 1

B) mv³/4√2 g

Maximum height reached by particle

H = u²sin²Ф/2g = v²/4g

Angular momentum

L = m (v horizontal component) H

L = m vcosФ H

L = m v³/4√2 g