Please explain straight line motion question ?

A tennis ball with a velocity of  +10.0 m/s to the right is thrown perpendicular at a wall.  After striking the wall, the ball rebounds in the opposite direction with a velocity of -8.0 m/s to the left.  If the ball is in contact with the wall for 0.012s, what is the average acceleration of the ball while it is in contact with the wall ?


This is High School Physics question from straight line motion under constant acceleration chapter.

Please explain the question.   Person standing in front of wall throws the ball towards the ball at +10.0 m/s.  The ball hits the wall and rebounds toward the person at -8.0 m/s.  Did I understand the problem correctly ?
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naseeamAsked:
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aburrConnect With a Mentor Commented:
The question makes sense only if the action takes place somewhere where there is no gravity.
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aburrConnect With a Mentor Commented:
In that case you can consider just the rebound.
The ball changes momentum. Can you calculate that change?
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aburrCommented:
actually you do not meed to calculate the momentum.
You know the velocity before and after the collision, you know the time it took to make that change in velocity.
Do you know the equation for average acceleration ?  

 If so write it. If not I can give it to you.
You know everything in that equation.
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ozoConnect With a Mentor Commented:
ball with a velocity of  +10.0 m/s to the right is thrown perpendicular at a wall.  After striking the wall, the ball rebounds in the opposite direction with a velocity of -8.0 m/s to the left.
I would have called it +8.0 m/s to the left, or  -8.0 m/s to the right.
But in any case, it amounts to a change of |18 m/s|
It looks like you did understand the problem correctly.
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aadihConnect With a Mentor Commented:
(-) 18/0.012 m/square-second.

[Change in velocity/Time.]

(-) = Deceleration.
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