Answer:
The magnitude of the average acceleration of the ball during this time interval is 10,449.44 m/s²
Explanation:
Given;
mass of super ball, m = 50 g
initial velocity of the ball, u = 25.5 m/s
final velocity of the ball, v = -21 m/s (re-bouncing backward)
time in contact with the wall, t = 4.45 ms = 0.00445 s
The average acceleration of the ball during this time interval is given by
[tex]a = \frac{dv}{dt} = \frac{v-u}{t} \\\\a = \frac{-21-25.5}{0.00445}\\\\a = -10449.44 \ m/s^2\\\\|a| = 10,449.44 \ m/s^2[/tex]
Therefore, the magnitude of the average acceleration of the ball during this time interval is 10,449.44 m/s²
