SOLUTION: Aidan drops a ball vertically from a height of 120 feet. The peak height after each bounce is half the previous height. How far does the ball travel from the time he drops it until

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Question 1153357: Aidan drops a ball vertically from a height of 120 feet. The peak height after each bounce is half the previous height. How far does the ball travel from the time he drops it until it reaches the peak height after the 4th bounce? Please explain exactly how you came up with your answer!
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Refer to the diagram below.

We start at a height of 120 feet. Then the ball is dropped to bounce back up again to a peak height of 60 feet (half that of 120). The diagram shows this with the use of arrows. Blue arrows point down while red point up. I've also marked in a lighter blue color each time there is a bounce point. This process keeps going until we reach the green point in the diagram, which is after the 4th bounce. The height of this green point is 7.5 feet.

We then add up the lengths of each arrow in the diagram
120 + 60+60 + 30+30 + 15+15 + 7.5 = 337.5

Answer: 337.5 feet

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Alternative Method

Refer to the diagram above. The sequence of blue arrow lengths
120, 60, 30, 15
is a geometric sequence with a = 120 as the first term and r = 0.5 as the common ratio

Compute the partial sum S4
Sn = a*(1-r^n)/(1-r)
S4 = 120*(1-0.5^4)/(1-0.5)
S4 = 225
The blue arrows have their lengths add to 225 feet.
As confirmation: 120+60+30+15 = 225


Do the same for the sequence: 60, 30, 15, 7.5
This is the lengths of the red arrows
This time, a = 60 is the first term. The common ratio stays the same.
Sn = a*(1-r^n)/(1-r)
S4 = 60*(1-0.5^4)/(1-0.5)
S4 = 112.5
The red arrows have their lengths add to 112.5 feet.
As confirmation: 60+30+15+7.5 = 112.5

Now add the two subtotals
225+112.5 = 337.5

We get the same answer of 337.5 feet