Answer:
The magnitude of the average acceleration of the ball during this time interval is 381.50 [tex]\frac{m}{s^{2} }[/tex]
Explanation:
1 ms = [tex]10^{-3}s[/tex]
Assuming a coordinate system:
v0 = -29 m/s
v1 = 19.5 m/s (rebounds is positive)
a = Δv/Δt = [tex]\frac{v1 - v0}{Delta t}[/tex] = [tex]\frac{19.5 - (-29.0) \frac{m}{s} }{4.65 ms * \frac{1s}{10^{3}ms } }[/tex] = [tex]\frac{48.5 \frac{m}{s} }{4.65 * 10^{-3}s }[/tex] = 381.50 [tex]\frac{m}{s^{2} }[/tex]
