A ball is dropped from a height of 1m. The ball bounces and rebounds, losing 14% of its previous height and then falls again. If the ball continues to rebound and fall in this manner, find the total vertical distance the ball travels until it hits the ground for the 6th time.

9514 1404 393

  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.

Let's give them the values Heads=0 and Tails=1 and we have a Random Variable "X":

In short:

X = {0, 1}

Note: We could choose Heads=100 and Tails=150 or other values if we want! It is our choice."

                   (  https://www.mathsisfun.com/data/random-variables-continuous.html  )

Sorry really busy but i hope this helps.