A particle of mass `m` is projected with a speed `v` at an angle `theta` with the horizontal. Find the angular momentum of the particle about an axis passing through point of projection and perpendicular to the plane of motion of the particle.
`(a)` When the particle is at maximum height and
`(b)` When the particle is just about to collide with the horizontal surface

Explanation

`(a)` Position vector of the particle when it is at maximum height `H` is
`vecr=xhati+Hhatj`
Linear momentum of the particle at maximum height
`vecp=mv cos theta hati`
therefore `vecL=vecrxxvecp`
`=(xhati+Hhatj)xxmv cos theta hati`
`=mvH cos theta(-hatk)`
`=mv cos theta(v^(2)sin^(2)theta)/(2g)(-hatk)`
`=(mv^(3)sin^(2)thetacostheta)/(2g)(-hatk)`
`(b)` Position vector of the particle just before hitting ground s
`vecr=Rhati`
`=(v^(2)sin2theta)/(g)hati`
Linear momentum of the particle is
`p=mv costheta hati-mv sin theta hatj`
Therefore `vecL=vecrxxvecp`
`=Rhatixx(mv cos theta hati-mv sin thetahatj)`
`=mvRsintheta(-hatk)`
`=(mvsinthetav^(2)sin2theta)/(g)(-hatk)`
`impliesvecL=(mv^(3)sin2thetasintheta)/(g)(-hatk)`