Answer:
The magnitude of the average acceleration of the ball during this time interval is 1.238 x 10⁴ m/s².
Explanation:
Given;
mass of the super ball, m = 50 g = 0.05 kg
initial velocity of the ball, u = 29.5 m/s
final velocity of the ball, v = -20.0 m/s (negative because it rebounds)
time of contact of the ball and the wall, t = 4 ms = 4 x 10⁻³ s
The force exerted on the brick wall by the ball is given as;
[tex]F = ma\\\\ma = \frac{m(v-u)}{t} \\\\a = \frac{v-u}{t} \\\\a = \frac{(-20) - 29.5}{4.0 \ \times \ 10^{-3}} \\\\a = \frac{-49.5}{4.0 \ \times \ 10^{-3}} \\\\a = -1.238 \times 10^4 \ m/s^2\\\\|a| = 1.238 \times 10^4 \ m/s^2[/tex]
Therefore, the magnitude of the average acceleration of the ball during this time interval is 1.238 x 10⁴ m/s².
