Superball sum of sequence question?

Question

"Superballs" bounce to 90% of the height from which they are dropped. If a "superball" is dropped from a height of 2 meters, how far does it travel...
a) when it hits the ground for the 10th time?
b) when it comes to rest?

Answer 1

Let the first height it dropped from is a
it will be rflected for the first time to a height = 0.9 a = a r
So it traveles till the 10th time a distanc d10 where :
d10 = a + 2a r + 2 a r^2 + ..... + 2 a r^9
d10 = 2 ( a + a r + a r^2 + ..... + a r^9) - a
d10 = 2 [ a(1 - r ^10)/(1-r)] = 2[ 2(1-(0.9)^10)/(1 - 0.9) ] -2
d10 = 11.0264312 meters
b) Till it comes to rest :
d infinty = 2( sum of the geometric sequence till infint ) - 2
= 2 [2/(1- 0.9)] -2 = 38 meters

Answer 2

A) 2 x .9^10 = .697358802 meters

B) x is the number of times the ball bounces. For every x, it's .9 of the last x. You start at 2. I'm not quite sure how to set up the equation, but you'd stop when y=0. Theoretically, the ball would never come to a rest, but that's not practical, because the ball's bounce percentage actually gets lower the farther down from the optimum height (height from which you can drop the ball and it will bounce to it's highest percentage) you go, eventually reaching 0%.

Answer 3

After the first bounce, it returns to .9 * 2m = 1.8m
After the tenth bounce it returns to
.9 * .9 * .9 * .9 * .9 * .9 * .9 * .9 * .9 * .9 * 2m = .69 m

After it comes to rest it doesn't go anywhere.