Elastic collision with the floor

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SUMMARY

The discussion centers on calculating the height a ball rebounds after an elastic collision with the floor. Given a mass of 100g and an initial drop height of 5.2m, the velocity just before impact is determined to be 10m/s. The consensus is that, in an ideal elastic collision without air resistance, the ball will rebound to the same height of 5.2m. Additionally, the impulse and average force can be calculated using momentum principles without needing the rebound height.

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Homework Statement


A ball of mass m = 100g , falls at an height of 5,2m and it collides elastically with the floor and comes up again. Find the height h2 (how much the ball has come up), the impulse, the average force if the collision lasts 0,02seconds.


Homework Equations





The Attempt at a Solution



I got the velocity just before the collision:
mgh = 1/2mv^2 => v = square root of (2gh) => v = 10m/s.
Okay, now I don't understand how I can find the height. If it was a completely elastic collision it'd still be 5,2 meters, but I guess it's not because it is actually asking to find it.

About the impulse and the average force I know how to find them, but I'd need the height first.
Thanks.
 
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From the wording of this question I would agree with you. 'Elastically' means KE before = KE after and this means the ball should go to the same height as long as air resistance can be ignored. There is no mention of air resistance so... I agree with you.
You do not need the height to calculate impulse and average force. You need to use ideas about momentum for that.
 
Okay, sounds good to me. Thanks.
 

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