Angular Moment

Relevant wiki: Relating Angular Momentum and Linear Momentum

Now, we know that $\vec{L} = \vec{r} \times \vec{p}$ , where $\vec{p}$ is the ball's linear momentum and $\vec{r}$ is the position vector of the ball from the point $P$ . In scalar form, the angular momentum, $L = m v r \sin \theta$ .

Let after time $t$ the angle between the velocity of the ball and its position vector from the point $P$ be $\theta$ , then $r_{\perp} = r \sin (180 - \theta) = r \sin \theta$ .

Hence, the angular momentum is $L = m v r_{\perp}$

Here $r_{\perp}$ is the perpendicular distance between the line of the velocity of the ball and the point $P$ to the building. Since both the point $P$ and the building are stationary, $r_{\perp}$ is constant.

Now, as the falling ball has an acceleration $g$ we can write the velocity of the ball $v$ after time $t$ as $v = gt$ .

Combining these results, we get $L = mgr_{\perp}t$ . As $r_{\perp}$ is constant, so $\boxed{L \propto t}$ .