A particle of mass `m` has been thrown with intial speed `u` making angle `theta` with the horizontal ground. Find the angular momentum of the projectile about an axis perpendicular to the plane and passing through the point of projection when the projectile is
`(a)` At the highest point
`(b)` About to hit the ground

Explanation

(a)
At the highest point the projectile has the momentum mu cos `theta` in horizontal direction. Now, extend a line backward from the momentum direction and draw a perpendicular `r_(_|_)` mu cos `theta = (H_("max"))` mu cos `theta`
`=(u^(2)sin^(2)theta)/(2g)(m u cos theta) " " (because H_("max")=(u^(2)sin^(2)theta)/(2g))`
`= (m u^(3)sin^(2)theta cos theta)/(2g)`
(b) When the projectile is about to hit the ground the situation is as follows. Body will have momentum mu at angle `theta` with the horizontal just before hitting the ground.

Now extend a line backward from momentum mu and draw a perpendicular. `r_(_|_)` on it form O. You can see `r_(_|_)=` (Range) `sin theta`
So, angular momentum about O is equal to
`r_(_|_)` mu = (Range) `sin theta m u=((u^(2)sin 2theta)/(g))sin theta mu " " (because " Range" = (u^(2)sin 2theta)/(g))`
`= ((u^(2)2 sin theta cos theta)/(g))sin theta. m u " " (because sin 2 theta = 2 sin theta cos theta)`
`= (2 m u^(3)sin^(2)theta cos theta)/(g)`.