SOLUTION
The ball is dropped from a height of h feet and repeatedly bounces of the floor
Since it reaches 2/3 of the height, we have
[tex]\begin{gathered} h\text{ initial height dropped + } \\ 2h\times\frac{2}{3}\text{ that is up and down with }\frac{2}{3}\text{ of h} \end{gathered}[/tex]The second bounce becomes
[tex]\begin{gathered} 2(\frac{2}{3})h \\ \\ \end{gathered}[/tex]So the bounces follow the series
[tex]h+2(\frac{2}{3})h+2(\frac{2}{3})^2h+2(\frac{2}{3})^3h+2(\frac{2}{3})^4h+2(\frac{2}{3})^5h[/tex]where a the first term = h
and r the common ratio = 2/3
So the total number of feet the ball travels between the first and sixth bounce is
[tex]\sum_{i\mathop{=}1}^5(2h)(\frac{2}{3})^i[/tex]Hence the answer is option A
