9514 1404 393
Answer:
7.51 m
Step-by-step explanation:
The distance traveled to the first floor hit is 1 m. The next hit has a height that is 14% less, so is 100% -14% = 86% = 0.86 times this height. The travel up and down to/from that height is added to the total.
The distance traveled to the second floor hit is 1 m + 2×(0.86×1 m).
The distance traveled to the third floor hit is 1 m + 0.86×2 m + 0.86²×2 m.
Then the distance traveled to the n-th floor hit is ...
-1 m + Sum(0.86^k × 2 m) for k = 0 to n-1
The sum can be found from the formula for the sum of a geometric series:
Sn = a1×(r^n -1)/(r-1) . . . . . where a1 is the first term and r is the common ratio.
For r = 0.86, n = 6, and a1 = (2m), the total distance is ...
vertical distance = -1 m + (2 m)(0.86^6 -1)/(0.86 -1)
= -1 m + (2 m)(-0.5954)/(-0.14) ≈ 7.5062 m
The total distance traveled to the 6th floor hit is about 7.51 meters.
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Alternate solution
Another way to answer the question is to add up the distances for each bounce. Attached is a spreadsheet that does that.
