Physics - W.E.P. Question with Solution | TestHub

After each collision speed becomes$e$times

$\therefore$Speed after$1^{st}$collision is${\nu }_1=e{\nu }_0$

$\therefore$Maximum height after$1^{st}$collision is,

$h_1=\frac{{\nu }_1^2}{2g}=e^2\left(\frac{{\nu }_0^2}{2g}\right)=e^{2}h$

Similarly maximum height after$2^{nd}$collision is,

$h_2=e^2{h}_1=e^{4}h$

maximum height after$3^{rd}$collision is,

$h_3=e^2{h}_2=e^{6}h$

$\therefore$The height to which the ball goes up after it rebounds for the$n^{th}$time is$h_n=e^{2n}h$