Question

A ball is dropped from a height of $48$ ft. and rebounds two-third of the distance it falls. If it continues to fall and rebound in this way, Let the total distance it will travel before coming to rest be $k$ft.Find $\frac{k}{48}$ ?

Answer 1

Let the total distance be $h=48$ feet.
In the first fall it covers a distance of
$h+\frac{2h}{3}$ ....(i)
In the second fall it covers a distance of
$\frac{2h}{3}+\frac{4h}{9}$ .....(ii)
In the third fall it covers a distance of
$\frac{4h}{9}+\frac{8h}{27}$ .....(iii) and so on.
Hence, the series is
$h+\frac{2h}{3}+\frac{2h}{3}+\frac{4h}{9}+\frac{4h}{9}+\frac{8h}{27}...\infty$
$=h+2[\frac{2h}{3}+\frac{4h}{9}+\frac{8h}{27}+...\infty ]$
$=h+2h[\frac{2}{3}+\frac{4}{9}+\frac{8}{27}+...\infty ]$
$=h+2h\left(\frac{\frac{2}{3}}{1-\frac{2}{3}}\right)$
$=h+2h\left(2\right)$
$=4h+h$
$=5h$
$=5\times 48$ feet
Hence, $k=5\times 48$ feet
Therefore, $\frac{k}{48}$
$=\frac{5\times 48ft}{48ft}$
$=5$