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A particle falls from a height 'h' upon a fixed horizontal plane and rebounds. If e is the coefficient of restitution, the total distance travelled before it stops rebounding is

\[\begin {array} {1 1} (a)\;h \bigg[ \frac{1+e^2}{1-e^2}\bigg] & \quad (b)\;h \bigg[ \frac{1-e^2}{1+e^2}\bigg] \\ (c)\;\frac{h}{2} \bigg[ \frac{1-e^2}{1+e^2}\bigg] & \quad  (d)\;\frac{h}{2} \bigg[ \frac{1+e^2}{1-e^2}\bigg] \end {array}\]

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the ball bounds with velocity ev
$h_1=\large\frac{e^2v^2}{2g}$
$h_2=\large\frac{[e^4v^2]}{2g}$
Total distance :
$h+2h_1+2h_2+2 h_3+..........$
$\large\frac{v^2}{2g}$$+2\large\frac{e^2v^2}{2g} $$+2 \bigg(\large\frac{e^4v^2}{2g}\bigg)$$+2 \large\frac{(e^6v^2)}{2g}$$+.............$
$\large\frac{v^2}{2g}+\frac{e^2v^2}{2g}$$\bigg[1+e^2+e^4+...... \bigg]$
$\large\frac{v^2}{2g}+\frac{e^2v^2}{g}\bigg[\frac{1}{1-e^2}\bigg]$
$\large\frac{[1-e^2+2e^2]v^2}{2g(1-e^2)}=\bigg[\frac{1+e^2}{1-e^2}\bigg]\frac{v^2}{2g}$
$\qquad= \bigg( \frac{1+e^2}{1-e^2}\bigg)$$h$
answered Nov 23, 2013 by meena.p
edited Jun 18, 2014 by lmohan717
 

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