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A ball is dropped from height h on the ground level. If the coefficient of restitution is e then the height upto which the ball will go after \(\sf n^{th}\) jump will be
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@IrishBoy123
@radar
@iambatman
\[v= \sqrt{2gh}\] \[e = \frac{ v' }{ v _{0} } = \sqrt{\frac{ H }{ h }}\] thus \[H = e ^{2}*h\] let's us now use the new height to make the same calculation \[H _{1}= e ^{2}*H = e ^{2}*e ^{2}*h = e ^{4}*h\] that is enough to get the pattern \[H _{n} = e ^{2*n} * h\]
Thanks @catch.me
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