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A tennis ball with a speed of 11.4 m/s is thrown perpendicularly at a wall. After striking the wall, the ball rebounds in the opposite direction with a speed of 8.64 m/s. If the ball is in contact with the wall for 0.0104 s, what is the average magnitude of acceleration of the ball while it is in contact with the wall?
a = (-8.64-11.4)/0.0104 = 1926 m/s2
To find the average magnitude of acceleration of the ball while it is in contact with the wall, we can use the average acceleration formula:
Average acceleration = change in velocity / time interval
First, let's find the change in velocity. The ball's initial velocity is 11.4 m/s, and its final velocity is -8.64 m/s (since it rebounds in the opposite direction). Therefore, the change in velocity is:
Change in velocity = final velocity - initial velocity
Change in velocity = -8.64 m/s - 11.4 m/s
Change in velocity = -20.04 m/s
Next, we need to calculate the time interval. The ball is in contact with the wall for 0.0104 s.
Now, we can plug these values into the formula to find the average magnitude of acceleration:
Average acceleration = change in velocity / time interval
Average acceleration = -20.04 m/s / 0.0104 s
Average acceleration = -1926.9231 m/s²
Since we are looking for the magnitude of acceleration, we ignore the negative sign. Therefore, the average magnitude of acceleration of the ball while it is in contact with the wall is approximately 1926.9231 m/s².