Answer
The angular momentum about $P$ is $~~-22.6~kg~m^2/s$
Work Step by Step
We can find the maximum height $h$:
$v^2 = v_0^2+2ah$
$h = \frac{v^2 - v_0^2}{2a}$
$h = \frac{0 - (40.0~m/s)^2}{(2)(-9.8~m/s^2)}$
$h = 81.63~m$
Halfway back down to the ground, the height is $40.815~m$
We can let $v_0 = 0$ since the ball falls from maximum height.
We can find the speed $v$ at this point:
$v^2 = v_0^2+2ay$
$v^2 = 0+2ay$
$v = \sqrt{2ay}$
$v = \sqrt{(2)(9.8~m/s^2)(40.815~m)}$
$v = 28.3~m/s$
Note that by the right hand rule, the angular momentum is negative.
We can find the angular momentum about $P$:
$L = r_{\perp}~mv$
$L = -(2.0~m)(0.400~kg)(28.3~m/s)$
$L = -22.6~kg~m^2/s$
The angular momentum about $P$ is $~~-22.6~kg~m^2/s$
