A particle of mass `m` is thrown with speed `v_(0)` at angle `30^(@)` with horizontal. Find the angular momentum about the point of projection when the particle is at the highest point of its trajectory.

Explanation

Angular momentum about `O`
`=` linear momentum `xxbot^(ar)` distance from `O`
`=mv_(0)cos30^(@)xxH_(max)`
`= mv_(0) cos30^(@)xx(v^(2)sin^(2)30^(@))/(2g)`
`=mv_(0)(sqrt3)/(2)xx(v_(0)^(2)xx(1//4))/(2g)`
`=(sqrt3mv_(0)^(3))/(16g)`