Answer
The tangential speed at a point on the equator of the ball is $4.64m/s$
Work Step by Step
The ball has a horizontal linear speed $v_x=42.5m/s$ and travels a distance of $16.5m$. The time the ball travels before being caught is $$t=\frac{16.5m}{42.5m/s}=0.388s$$
The ball rotates through an angle of $49\ rad$. Its angular speed is $$\omega=\frac{49rad}{0.388s}=126.3rad/s$$
The baseball's radius $r=3.67cm=3.67\times10^{-2}m$. The tangential speed at a point on the equator is $$v_T=\omega r=4.64m/s$$
