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Let's say a ball is dropped from the height h on the ground and bounces repeatedly. If the coefficient of restitution is e, the height to which the ball goes up after it rebounds for the $nth$ time is H = $he^{2n}$ Now if we find the maximum height of the ball between the $nth$ and $(n+1)th$ impact with the ground, it should be $H = he^{2n+2}$ but my textbook says is it $H = he^{2n}$. How? The expression for the height for the nth time and between the nth and (n+1)the time will be the same?

Is it because for the maximum height between nth and (n+t)th, the ball is bouncing nth time so the answer becomes $H = he^{2n}$ and the ball didn't bounced for the (n+1)th time?

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  • $\begingroup$ Are you sure that it's $H=he^{2n}$, not $H=he^{-2n}$ ? Height should decrease after $n$ hits. $\endgroup$
    – Agnius Vasiliauskas
    Commented Jun 17 at 5:57
  • $\begingroup$ @AgniusVasiliauskas No, it's no negative $\endgroup$
    – Moksh Singh Dangi
    Commented Jun 19 at 13:40

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The height after the 1st bounce: 𝐻1= β„Žπ‘’^2

The height after the 2nd bounce: 𝐻2=β„Že^4

And so on, for the nth bounce: 𝐻𝑛=β„Žπ‘’^(2𝑛)

However, if we are looking at the maximum height between the nth and (n+1)th bounce, this height would be achieved immediately after the nth bounce (before the ball loses more energy in the next impact). This maximum height is what the ball reaches after bouncing off the ground for the nth time. Therefore, between the nth and (n+1)th bounce, the maximum height is indeed 𝐻𝑛=β„Žπ‘’^(2𝑛)

The expression: 𝐻𝑛+1=β„Žπ‘’^(2𝑛+1) would represent the maximum height reached after the ball bounces again and hits the ground for the (n+1)th time, not the maximum height between the two bounces.

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    $\begingroup$ Thank you soo much. I was very confused. $\endgroup$
    – Moksh Singh Dangi
    Commented Jun 17 at 4:30
  • $\begingroup$ If you enclose your equations with $ (so one at the beginning of the equation, and one at the end), it will display them nicely (in LaTeX) $\endgroup$
    – Syrocco
    Commented Jun 17 at 6:28

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