Answer
The angular displacement of the ball is $11.7rad$
Work Step by Step
The ball leaves the edge with vertical velocity $v_0=0$, $g=9.8m/s^2$, and falls for a height $h=2.1m$. The amount of time the ball is in the air is $$h=v_0t+\frac{1}{2}gt^2$$ $$t=\sqrt{\frac{2h}{g}}=0.65s$$
We know the ball's linear velocity $v=3.6m/s$ and radius $r=0.2m$. Therefore, $$\omega=\frac{v}{r}=18rad/s$$
which is the angular speed of the ball.
Therefore, the angular displacement of the ball is $\theta=\omega t=11.7rad$
