Question

A challenge Question
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Note: Don't misunderstand that I need help....Just for fun....Try to explain the best & then I will do...

A body of mass m falls from height h on ground. If 'e' be the coefficient of restitution of collision between the body and ground, then the distance it travels before coming to rest

Answer 1

you build a geometric series in e

copying from Wiki: \(\text{Speed of separation} = e \times \text{Speed of approach}\)

so if hits deck first at velocity \(v \{= \sqrt{2gh} \}\), then it rebound at \(e v\). It rises to height \(\dfrac{(ev)^2}{2g}\) and so covers a distance twice that before it lands again, ie \(\dfrac{(ev)^2}{g}\)

then rebounds at speed \(e^2v\), so covers total distance for next cycle of \(\dfrac{(e^2v)^2}{g}\)

and so on

so we are adding \(h + \dfrac{e^2v^2}{g} + \dfrac{e^4v^2}{g} + \dots\)

or \(h+ 2he^2 + 2h e^4 + \dots \)
\(= 2h+ 2he^2 + 2h e^4 + \dots -h\)
\(= 2h(1+ e^2 + e^4 + \dots) -h)\)
\(=\dfrac{2h}{1-e^2}-h\) or other algebraic equiv.