Finding the height of a bouncing ball. (Using the geometric series in an applied setting)

Question

I am doing a problem in a textbook (Boas' Mathematical Methods in the Physical Sciences) where a ball is dropped from a height of one yard and the sum of vertical distance in each drop is the series: $1+2\cdot\dfrac{2}{3}+2\cdot\dfrac{4}{9}+2\cdot\dfrac{8}{27}+\cdot \cdot \cdot=1+2\left(\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{8}{27}+\cdot\cdot\cdot(\dfrac{2}{3})^n\right)$. I am asked to find the height of the tenth rebound and the total vertical distance traveled when the ball touches the ground for the tenth time.

I know that the height of the tenth bounce is just the tenth term of the sequence so $(\dfrac{2}{3})^{10}\approx0.0173$ yards. Now I thought that the vertical distance would be $1+2\cdot\dfrac{\frac{2}{3}(1-(\frac{2}{3})^{10})}{1-\frac{2}{3}}\approx1.96$ yards using the formula for a geometric series $\dfrac{a(1-r^{n})}{1-r}$. However, the book tells me that the answer should be $6\cdot(\dfrac{2}{3})^{10}\approx0.104$ yards. Now on the chance that I did misinterpret the book and the author meant the vertical distance the ball traveled the tenth time it hits the ground, shouldn't it be $2\cdot\dfrac{2}{3}^{10}$?

Answer 1

When the ball touched the ground for the first time, it travelled 1 yard.

On the second time, $1 + 2 \cdot(\frac23)^1$ yards.

So on the tenth time, $1 + 2 \cdot (\frac23)^1+\ldots+2\cdot(\frac23)^9=1+2\cdot\dfrac{\frac23(1-(\frac{2}{3})^{9})}{1-\frac23} \approx 4.90$ yards.

And if the question were the distance travelled after the ninth time it touched the ground and just before the tenth, then it is $(\frac{2}{3})^{9}$ yards.

Answer 2

Let the vertical distance traveled in each drop(excluding the first drop, as there is only comming down) be $d_n=$(going up+comming down)

The vertical distance traveled in the 10th drop,$d_{10}=2.\left(\frac{2}{3}\right)^{10-1}$

Height of the 10th drop is $\left(\frac{2}{3}\right)^9$ (distance traveled up is same as distance traveled down)

$1+2\cdot\frac{2}{3}+2\cdot\frac{4}{9}+2\cdot\frac{8}{27}+...=1+2\left(\frac{2}{3}+\frac{4}{9}+\frac{8}{27}...+ \left(\frac{2}{3}\right)^9\right)$

bracket quantity is GP with $a=\frac{2}{3}, r=\frac{2}{3}, n=9$

Answer 3

Answer:

Going by the answer given by Mark Leo,

Part B) $= 1+2\cdot\dfrac{\frac23(1-(\frac{2}{3})^{9})}{1-\frac23}$

If you simplify it like below $$= 1+2\times2\left[1-(\frac23)^{10}\right]$$

Further simplifying, you get $$ 5-4\times(\frac23)^{9} = 5-2\times3\times(\frac23)^{10} = 5-6\times(\frac23)^{10}$$

You must be in the turn of page and are missing (5 -) in the answer. All these things happen in printed books. Good luck

Thanks

Satish