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Why is the speed of rebound the same as the speed of approach for the second collision?
Because there is no way for the ball to lose energy whilst in the air - there is no air resistance.

Alternatively, on the way up, the ball starts with speed v upwards, travels a distance h, say, decelerating at g m/s^2. On the way down, the ball starts with speed 0, travels a distance h, accelerating at g m/s^2. By SUVAT, the final speed on the way down musy again have magnitude v.
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Zenarthra
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Original post by atsruser
Because there is no way for the ball to lose energy whilst in the air - there is no air resistance.

Alternatively, on the way up, the ball starts with speed v upwards, travels a distance h, say, decelerating at g m/s^2. On the way down, the ball starts with speed 0, travels a distance h, accelerating at g m/s^2. By SUVAT, the final speed on the way down musy again have magnitude v.


Thanks!

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