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A 50.0-g Super Ball traveling at 29.5 m/s bounces off a brick wall and rebounds at 19.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 4.75 ms, what is the magnitude of the average acceleration of the ball during this time interval

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Answer:


a=-10210.52\ m/s^2

Step-by-step explanation:

Given that,

Mass of a ball, m = 50 g

It is traveling at 29.5 m/s bounces off a brick wall and rebounds at 19.0 m/s.

Initial velocity, u = 29.5 m/s

Finl velocity, v =-10 m/s (as it rebounds)

We need to find the magnitude of the average acceleration of the ball during this time interval.

Acceleration = rate of change of velocity


a=(v-u)/(t)\\\\a=((-19)-29.5)/(4.75* 10^(-3))\\\\=-10210.52\ m/s^2

So, the required acceleration is
10210.52\ m/s^2.

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