Question

A ball is dropped from a height of 900 cm. Each time it rebounds, it rises to 2/3 of the height it has fallen through. Find the two times of total distance travelled by the ball before it comes to rest in decameters.

Answer 1

According to question. total distance $=h+2\times \frac{2}{3}h+2\times (\frac{2}{3}{)}^{2}h+2\times (\frac{2}{3}{)}^{3}h+....upto\infty$
We can rewrite it as $h+2[\frac{2}{3}h+(\frac{2}{3}{)}^{2}h+.....upto\infty ]$ ........(1)

Here, $\frac{2}{3}h+(\frac{2}{3}{)}^{2}h+.....upto\infty$ is an infinite G.P. with $a=\frac{2}{3}h$ and $r=\frac{2}{3}$

Using the formula for sum of infinite terms of a G.P. $\left(\frac{a}{1-r}\right)$, we get,

$\frac{2}{3}h+(\frac{2}{3}{)}^{2}h+.....upto\infty =2h$

Putting that in eq.(1), we get,

$\left(h\right)+2\left[2h\right]=5h$

$\therefore$ two times the total distance $=2\times 5h=10h$

Given that, $h=900cm$

Hence, $10h=9000cmor9decameters$