SOLUTION: Please help solve this. It would help me a lot in reviewing for my final exam. Thank you so much! A rubber ball is dropped from a height of 12 feet. If it rebounds one-third of

The ball bounces to a 1/3 its initial height every time, which means the series of heights forms a geometric progression, with common ratio as 1/3

First term a = Initial height = 12

2nd term = 1st rebound = 12*1/3 = 4

3rd term = 2nd rebound = 4*1/3 = 4/3



So the height of the 8th rebound is given by the 9th term in the series.

9th term = a*r^(n-1) = 12*(1/3)^8 = 12/6561 = 4/2187



Height of rebound for 8th time = 4/2187.





2) Total distance travelled at the 10th time it hits ground is the sum of  the 

upward travel and the downward travels. Both are geometric progressions.



Downward travel:

1st term = 12

Common ratio = 1/3

Sum of 10 terms = 









= 



= 



2) Upward (rebound) travel



1st term = 4

Common ratio = 1/3

Sum of 9 terms = (since it rebounds 9 times)









= 



= 



So the total travel = 17.999 + 5.999 = 23.998 approximately



As you would see, the sum is already approaching the "sum to infinity" of the

series.





:)