SOLUTION: I've tried different ways to write an equation but am having no luck. Can anyone help? A tennis ball dropped onto a hard surface rebounds exactly 2/3 the height from which it f

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Question 177428: I've tried different ways to write an equation but am having no luck. Can anyone help?
A tennis ball dropped onto a hard surface rebounds exactly 2/3 the height from which it falls. How far will it rebound after hitting the surface for the third time if it is dropped from an initial height of 18 feet?
Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
After the first drop, it rebounds back to the height of %282%2F3%29%2818%29=36%2F3=12 feet. So it starts at 18 ft, drops to 0 ft, then bounces back to 12 ft


Now after the second drop, it bounces back to a height of %282%2F3%29%2812%29=24%2F3=8 ft. So the ball starts at 12 ft, goes to 0 ft, then back up to 8 ft


Finally, after the third drop, the ball rebounds to a height of about %282%2F3%29%288%29=16%2F3=5.333 (note: this is approximate) feet. So it starts at 8 ft, drops to 0 ft, and then back up to roughly 5.333 ft


So the answer is 5.333 ft


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Notice how I took each previous answer and I multiplied it by 2%2F3. So on the first drop, I multiplied 2%2F3 by 18 to get the following (Let's call that result "a")

So a=18%2A%282%2F3%29


Now I then multiplied that result by another 2%2F3 to get (now let's call it "b")

b=a%2A%282%2F3%29


b=18%2A%282%2F3%29%2A%282%2F3%29 Plug in a=18%2A%282%2F3%29



Finally, I multiplied that previous result "b" by 2%2F3 to get the next height (which we'll call "c")


c=b%2A%282%2F3%29


c=18%2A%282%2F3%29%2A%282%2F3%29%2A%282%2F3%29 Plug in b=18%2A%282%2F3%29%2A%282%2F3%29


So after the third drop, there are 3 2%2F3 terms. After the second drop, there are 2 2%2F3 terms. After the first drop, there's only one 2%2F3 term.


So this means that after "x" drops (where you can plug any positive number in for "x"), there will be "x" 2%2F3 terms after the initial "18". So the equation might look like this:





To make things look tidy, we can condense the "x" number of 2%2F3 terms to get

h=18%282%2F3%29%5Ex


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Answer:

So the equation is h=18%282%2F3%29%5Ex where "h" is the height and "x" is the number of drops.


Notice how if we let x=3, then


h=18%282%2F3%29%5E%283%29


h=18%288%2F9%29


h=2%288%2F3%29


h=16%2F3


h=5.333


which was our original answer.



Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
The first time it falls, it rebounds to %282%2F3%29h if h is the initial height. For the second time, h is no longer the initial height, %282%2F3%29h is, so the rebound height becomes %282%2F3%29%282%2F3%29h or %282%2F3%29%5E2h. In general, the rebound height for the nth bounce is %28%282%2F3%29%5En%29h. Therefore, you need to calculate %28%282%2F3%29%5E3%2918