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Question Number 50913 by peter frank last updated on 22/Dec/18

A ball is dropped from height h.it strikes the ground and rises,the coefficiate of restitution being e what is the total distance it moves and the time before it comes to rest?

$${A}\:{ball}\:{is}\:{dropped}\:{from} \\ $$$${height}\:{h}.{it}\:{strikes}\:{the}\:{ground} \\ $$$${and}\:{rises},{the}\:{coefficiate} \\ $$$${of}\:{restitution}\:{being}\:\:{e} \\ $$$${what}\:{is}\:{the}\:{total}\:{distance} \\ $$$${it}\:{moves}\:\:{and}\:\:{the}\:{time}\: \\ $$$${before}\:{it}\:{comes}\:{to}\:{rest}? \\ $$$$ \\ $$$$ \\ $$

Answered by mr W last updated on 22/Dec/18

h_0 =h=initial height energy of ball before collision is mgh_0 E_(before) =mgh_(before) =(1/2)mv_(before) ^2 after the collision, the speed of the ball is reduced to v_(after) =ev_(before) E_(after) =(1/2)mv_(after) ^2 =mgh_(after) ⇒(h_(after) /h_(before) )=((v_(after) /v_(before) ))^2 =e^2 ⇒h_(after) =e^2 h_(before) if the max. height of the ball before collision is h, its max. height after collision is then e^2 h. h_1 =e^2 h=max. height after first collision h_2 =e^2 h_1 =e^4 h=max. height after second collision ..... total distance covered by the ball is s=h_0 +2h_1 +2h_2 +.... s=−h_0 +2(h_0 +h_1 +h_2 +....)=−h+2h(1+e^2 +e^4 +....) =−h+((2h)/(1−e^2 ))=h(((1+e^2 )/(1−e^2 ))) assume a collision takes no time, we consider only the time of the ball in air: t_0 =(√((2h_0 )/g))=(√((2h)/g)) time before first collision t_1 =2(√((2h_1 )/g))=2e(√((2h)/g)) time between first and second collisions t_2 =2(√((2h_2 )/g))=2e^2 (√((2h)/g)) time between second and third collisions ..... total time is t=t_0 +t_1 +t_2 +.... =−(√((2h)/g))+2(√((2h)/g))(1+e+e^2 +....) =−(√((2h)/g))+2(√((2h)/g))((1/(1−e))) =(√((2h)/g))(((1+e)/(1−e)))

$${h}_{\mathrm{0}} ={h}={initial}\:{height} \\ $$$${energy}\:{of}\:{ball}\:{before}\:{collision}\:{is}\:{mgh}_{\mathrm{0}} \\ $$$${E}_{{before}} ={mgh}_{{before}} =\frac{\mathrm{1}}{\mathrm{2}}{mv}_{{before}} ^{\mathrm{2}} \\ $$$${after}\:{the}\:{collision},\:{the}\:{speed}\:{of}\:{the} \\ $$$${ball}\:{is}\:{reduced}\:{to}\:{v}_{{after}} ={ev}_{{before}} \\ $$$${E}_{{after}} =\frac{\mathrm{1}}{\mathrm{2}}{mv}_{{after}} ^{\mathrm{2}} ={mgh}_{{after}} \\ $$$$\Rightarrow\frac{{h}_{{after}} }{{h}_{{before}} }=\left(\frac{{v}_{{after}} }{{v}_{{before}} }\right)^{\mathrm{2}} ={e}^{\mathrm{2}} \\ $$$$\Rightarrow{h}_{{after}} ={e}^{\mathrm{2}} \:{h}_{{before}} \\ $$$${if}\:{the}\:{max}.\:{height}\:{of}\:{the}\:{ball}\:{before}\:{collision} \\ $$$${is}\:{h},\:{its}\:{max}.\:{height}\:{after}\:{collision}\:{is} \\ $$$${then}\:{e}^{\mathrm{2}} {h}. \\ $$$$ \\ $$$${h}_{\mathrm{1}} ={e}^{\mathrm{2}} {h}={max}.\:{height}\:{after}\:{first}\:{collision} \\ $$$${h}_{\mathrm{2}} ={e}^{\mathrm{2}} {h}_{\mathrm{1}} ={e}^{\mathrm{4}} {h}={max}.\:{height}\:{after}\:{second}\:{collision} \\ $$$$..... \\ $$$${total}\:{distance}\:{covered}\:{by}\:{the}\:{ball}\:{is} \\ $$$${s}={h}_{\mathrm{0}} +\mathrm{2}{h}_{\mathrm{1}} +\mathrm{2}{h}_{\mathrm{2}} +.... \\ $$$${s}=−{h}_{\mathrm{0}} +\mathrm{2}\left({h}_{\mathrm{0}} +{h}_{\mathrm{1}} +{h}_{\mathrm{2}} +....\right)=−{h}+\mathrm{2}{h}\left(\mathrm{1}+{e}^{\mathrm{2}} +{e}^{\mathrm{4}} +....\right) \\ $$$$=−{h}+\frac{\mathrm{2}{h}}{\mathrm{1}−{e}^{\mathrm{2}} }={h}\left(\frac{\mathrm{1}+{e}^{\mathrm{2}} }{\mathrm{1}−{e}^{\mathrm{2}} }\right) \\ $$$$ \\ $$$${assume}\:{a}\:{collision}\:{takes}\:{no}\:{time},\:{we} \\ $$$${consider}\:{only}\:{the}\:{time}\:{of}\:{the}\:{ball}\:{in} \\ $$$${air}: \\ $$$${t}_{\mathrm{0}} =\sqrt{\frac{\mathrm{2}{h}_{\mathrm{0}} }{{g}}}=\sqrt{\frac{\mathrm{2}{h}}{{g}}}\:{time}\:{before}\:{first}\:{collision} \\ $$$${t}_{\mathrm{1}} =\mathrm{2}\sqrt{\frac{\mathrm{2}{h}_{\mathrm{1}} }{{g}}}=\mathrm{2}{e}\sqrt{\frac{\mathrm{2}{h}}{{g}}}\:{time}\:{between}\:{first}\:{and}\:{second}\:{collisions} \\ $$$${t}_{\mathrm{2}} =\mathrm{2}\sqrt{\frac{\mathrm{2}{h}_{\mathrm{2}} }{{g}}}=\mathrm{2}{e}^{\mathrm{2}} \sqrt{\frac{\mathrm{2}{h}}{{g}}}\:{time}\:{between}\:{second}\:{and}\:{third}\:{collisions} \\ $$$$..... \\ $$$${total}\:{time}\:{is} \\ $$$${t}={t}_{\mathrm{0}} +{t}_{\mathrm{1}} +{t}_{\mathrm{2}} +.... \\ $$$$=−\sqrt{\frac{\mathrm{2}{h}}{{g}}}+\mathrm{2}\sqrt{\frac{\mathrm{2}{h}}{{g}}}\left(\mathrm{1}+{e}+{e}^{\mathrm{2}} +....\right) \\ $$$$=−\sqrt{\frac{\mathrm{2}{h}}{{g}}}+\mathrm{2}\sqrt{\frac{\mathrm{2}{h}}{{g}}}\left(\frac{\mathrm{1}}{\mathrm{1}−{e}}\right) \\ $$$$=\sqrt{\frac{\mathrm{2}{h}}{{g}}}\left(\frac{\mathrm{1}+{e}}{\mathrm{1}−{e}}\right) \\ $$

Commented by peter frank last updated on 22/Dec/18

sorry sir more explanation with diagram if possible

$${sorry}\:{sir}\:{more}\:{explanation}\:{with}\: \\ $$$${diagram}\:{if}\:{possible} \\ $$

Commented by peter frank last updated on 22/Dec/18

the ans given is total distance=((h+he^2 )/(1−e^2 )) total Time=(√((2h)/g)) [((1+e)/(1−e))]

$${the}\:{ans}\:{given}\:{is} \\ $$$${total}\:{distance}=\frac{{h}+{he}^{\mathrm{2}} }{\mathrm{1}−{e}^{\mathrm{2}} } \\ $$$${total}\:{Time}=\sqrt{\frac{\mathrm{2}{h}}{{g}}}\:\left[\frac{\mathrm{1}+{e}}{\mathrm{1}−{e}}\right] \\ $$

Commented by peter frank last updated on 22/Dec/18

yes sir its true.thank you very much

$${yes}\:{sir}\:\:{its}\:{true}.{thank}\:{you} \\ $$$${very}\:{much} \\ $$

Commented by mr W last updated on 23/Dec/18

it′s interesting to know that the ball will infinitely move down and up, but all this happens within a limited time and limited distance.

$${it}'{s}\:{interesting}\:{to}\:{know}\:{that}\:{the}\:{ball} \\ $$$${will}\:{infinitely}\:{move}\:{down}\:{and}\:{up}, \\ $$$${but}\:{all}\:{this}\:{happens}\:{within}\:{a}\:{limited}\:{time} \\ $$$${and}\:{limited}\:{distance}. \\ $$

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