A particle of mass m is projected with velocity v at an angle `theta` with the horizontal. Find its angular momentum about the point of projection when it is at the highest point of its trajectory.

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)`

`rArr` Angular momentum `L=(mv^(3)"sin"^(2)theta cos theta)/(2g)`