## gr12 3 years ago On average, when a basketball is dropped, it bounces 7/10 as high as it did on the previous bounce. If a ball is dropped from a height of 10 m, determine the total distance up and down that the ball travels by top of the 11th bounce. Round to the nearest hundredth

1. theabsurd

$10(7\div10)^{11}$

2. dumbcow

it is the sum of a geometric sequence Sum = 10[(1-.7^11)/(1-.7)] = 32.67

3. Jemurray3

Are you familiar with the sum of a geometric series? Let $S = 1 + x + x^2 + x^3 +...+x^n$ then $xS = x + x^2 + x^3 + ...+x^{n+1}$ and so $S - xS = S(1-x) = 1 - x^{n+1}$ So $S = \frac{1-x^{n+1}}{1-x}$ That's the sum of a geometric series. Now examining your problem, we have to add the following: $10 + \frac{7}{10} \cdot 10 + \frac{7}{10}\cdot 10 + \left(\frac{7}{10}\right)^2 \cdot 10 + \left(\frac{7}{10}\right)^2 \cdot 10 + ...$ so on and so forth. Noting that the distance traveled (one-way, either up or down) after the nth bounce is $\left(\frac{7}{10}\right)^n \cdot 10$ and that everything except the first and last bounce is repeated, we can say that our total distance traveled is $\left[2\sum_{n=0}^{11} \left(\frac{7}{10}\right)^n\cdot 10\right]- 10 - \left(\frac{7}{10}\right)^{11}\cdot 10$ but this looks just like a geometric series with x = 7/10, so that means $\sum_{n=0}^{11} \left(\frac{7}{10}\right)^n = \frac{1-\left(\frac{7}{10}\right)^{12}}{1-\frac{7}{10}}$ And the rest is just calculation... :)