Answer:
18.33 Ns
Explanation:
As the pitch back speed has the opposite direction as before, the change in velocity would be
[tex]\Delta v = v_2 - v_1 = 29 - (-10) = 39 m/s[/tex]
So the change in momentum of the ball would be the product of its velocity change and its mass
[tex] \Delta p = \Delta v m = 39 * 0.47 = 18.33 kgm/s[/tex]
This is equals to the impulse acted on the ball by the bat, which is 18.33 Ns