A Rebounding Problem

Each bounce is two thirds of previous height:

${ h }_{ n }=\frac { 2 }{ 3 } { h }_{ n-1 }$

Total distance is the original height plus twice the height of each bounce:

$d=h+2\sum _{ n=1 }^{ \infty }{ { (\frac { 2 }{ 3 } ) }^{ n } } h$

Simplifying:

$d=h+2h\sum _{ n=1 }^{ \infty }{ { (\frac { 2 }{ 3 } ) }^{ n } }$

Sum to infinity:

${ S }_{ \infty }=\frac { a }{ 1-r }$ where $a=\frac { 2 }{ 3 } ,\quad r=\frac { 2 }{ 3 }$

${ S }_{ \infty }=\frac { 2 }{ 3 } \div \left( 1-\frac { 2 }{ 3 } \right) =\frac { 2 }{ 3 } \div \frac { 1 }{ 3 } =2$

Therefore:

$d=h+2h\times 2=h+4h=5h$