Question

A rubber ball dropped on a hard surface takes a sequence of bounces, each one 4/5 as high as the preceding one. If this ball is dropped from a height of 10 feet, what is the total vertical distance it has traveled after it hits the surface the fifth time? Be sure to account for both drop and rebound distances. We are learning sequences and series I'm confused on how to do this >.<

Answer 1

Before the first bounce: 10 ft to the ground
after the first bounce: 10(4/5) back up, which then has to go back down again, so:

10(4/5)(2) = 80/5 = 16

after the second bounce: 8(4/5) back up, which then has to go back down again, so:

8(4/5)(2) = 64/5

after the third bounce: (32/5)(4/5) back up, which then has to go back down, so:

(32/5)(4/5)(2) = 256/25

after the fourth bounce: (128/25)(4/5) back up, which then has to go back down, so:

(128/25)(4/5)(2) = 1024/125

Then the ball hits for the fifth time.

So now we add up the distances travelled.

10 + 16 + 64/5 + 256/25 + 1024/125
1250/125 + 2000/125 + 1600/125 + 1280/125 + 1024/125
7154/125
57 29/125 ft

Answer 2

You can do this using a G.P. as follows:
Consider the 2nd bounce: 10 ft x 4/5 = 8 ft.
For 4 bounces (note: we don't use 5 as we are not considering from the 1st bounce),
n = 4; T1=8; r=0.8
Sum of a G.P. = 8[1-(0.8^4)]/(1-0.8) = 23.616 ft.
Since we need to factor in the rebounds and the initial fall (10ft), we have:
Total vertical distance travelled = 2(23.616)+10 ft = 57.232 ft or 57 29/125 ft (Answer)