A particle of mass `m` is projected from the ground with an initial speed `u` at an angle `alpha`. Find the magnitude of its angular momentum at the highest point of its trajector about the point of projection.

Explanation

At the highest point in the only horizontal velocity
`v_(S)=vcostheta`
length of the perpendicular to the horizontal velocity from `O` is the maximum
height where
`H_("max")=(v^(2)sin^(2)theta)/(2g)`
`implies` Angular momentum
`L=(mv^(3)sin^(2)thetacostheta)/(2g)`