Soccer ball rebouncing of the goal post

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SUMMARY

The discussion focuses on calculating the average acceleration of a 420 g soccer ball that rebounds off a goal post. The initial velocity (v_i) is 23 m/s, and the final velocity (v_f) is -18 m/s, with a contact time of 3.3 ms. The correct formula for average acceleration is applied, resulting in an acceleration of approximately 12,121.21 m/s². The conversation emphasizes the importance of considering the change in velocity direction during the rebound and acknowledges the impulsive nature of the forces involved.

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Homework Statement


A 420 g soccer ball traveling at 23 m/s hits the goal post and rebounds at 18 m/s. If the ball is in contact with the post for 3.3 ms, what is the average acceleration of the ball during this time interval?

Homework Equations


Here I used 23m/s as initial velocity v_i and 18m/s as the final velocity v_f, and t = 0.0033s as the time traveled.

The Attempt at a Solution


I used a_x = v_f - v_i / t, so 18-23/0.0033 which gave me -1515.15m/s^2, which in my opinion seems to be a whole lot. Then I noticed that I can't use the goal post contact time as time traveled. So I am trying to find the time the ball actually traveled from start to hit the post and from rebound to the stop of the ball. Given those values I am stuck finding out how to get the time.
 
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Ah..you made a mistake..a=change in velocity/time taken
the direction of ball changes after rebound so change in velocity=(18-(-23))
And acceleration=(40)/0.0033
and also don't worry about the magnitude of acceleration.The force acting on the soccer will be impulsive(have very large value) so the acceleration will also be very high
 

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