Question 1117078: A ball from 10m above ground rebonds 3/4 of the distance from where it fell.
a) Find the total vertical distance the ball travels before coming to a rest. (Answer: 70m)
b) Find the total vertical distance the ball travels after 10 bounces. (Answer: 65.495)
c) What is the total distance traveled after the 10th bounce? (Answer: 4.505m)
I am not sure how to do them
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! This is a series with ratio (3/4) and can be considered a sum of 10+(3/4)10+(3/4)(7.5) etc. The sum in one direction, down, down, down, down is 40.
The sum in the other direction is up, up, up, but they are not equivalent with the first one, only after that.
The sum is 30 each way AFTER the first bounce. There is an additional 10 at the first bounce and that makes 70 m.
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Sum is a1(1-r^n)/(1-r)
=10(1-.75^10)/(.25)
=37.747 m
Multiply that by 2 (75.495 m) and subtract 10 for the fact it went down 10 m but didn't come up 10, and that is 65.495 m.
After the 10th bounce is the remainder of the distance to 70 m of 4.505 m.
The formula for how far it goes until the end is
a/(1-r), where a is the start and r is the ratio between each bounce
Draw it out crudely, don't worry about scale, and one can see that the total down distance is 40 (from the above formula) and the up distance has to be 30.
The sum formula computes the HEIGHT for each bounce and adds the heights only, not both up and down. This is why drawing what is going on is helpful.
drop
10
up 7.5
down 7.5, and the distance for bounce 2 only is twice that, or 15. Keep going, adding bounces 3-10
5.625
4.22
3.165
2.373
1.780
1.335
1.00
0.75
The sum of all of these DOUBLED (because the total distance is both ways) is 55.494 (EXCEPT the first drop). Add in the first drop and that is 65.494.
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