A body falls from a height of h = 3m. What is the total distance (in meters) it travels before it stops, if thebody falls from a height h, strikes the ground and rebounds to a height $e^{2} h$ , falls down $e^{2} h$ and then rebounds back to a height $e^{4} h$ and so on. The coefficient of restitution between body and ground is e = 0.55.

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Solution

Total distance travelled = $h + 2 e^{2} h + 2 e^{4} h + . . .$
$= h + 2 e^{2} h [1 + e^{2} + e^{4} . . . .] = h + 2 e^{2} h [\frac{1}{1 - e^{2}}] = h (\frac{1 + e^{2}}{1 - e^{2}}) = 3 [\frac{1 + (0.5)^{2}}{1 - (0.5)^{2}}]$ = 5m

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Total distance travelled = h+2e2h+2e4h+...=h+2e2h[1+e2+e4....]=h+2e2h[11−e2]=h(1+e21−e2)=3[1+(0.5)21−(0.5)2] = 5m

Total distance travelled = $h + 2 e^{2} h + 2 e^{4} h + . . .$
$= h + 2 e^{2} h [1 + e^{2} + e^{4} . . . .] = h + 2 e^{2} h [\frac{1}{1 - e^{2}}] = h (\frac{1 + e^{2}}{1 - e^{2}}) = 3 [\frac{1 + (0.5)^{2}}{1 - (0.5)^{2}}]$ = 5m