A particle of mass m is projected with a velocity `mu` at an angle of `theta` with horizontal.The angular momentum of the particle about the highest point of its trajectory is equal to :

Explanation

At the highest point it has only horizontal velocity
`v_(x) = v cos theta`

Length of the perpendicular to the horizontal velocity from O is the maximum height, where
`H_(max)=(v^(2) sin^(2)theta)/(2g)`
Angular momentum `= mvr = mv_(x) H_(max)`
`=mv_(x)((v^(2)sin^(2)theta)/(2g))`
`=(mv^(2) sin^(2) theta)/(2g)v_(x)`
`rArr` Angular momentum
`L=(mv^(3)sin^(2)thetacos theta)/(2g)`