SOLUTION: A rubber ball is dropped from the height of 13.5 feet. After each bounce the ball only goes up by 60% of what it did on the previous bounce. How high will the ball go after the 38t

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Question 219570: A rubber ball is dropped from the height of 13.5 feet. After each bounce the ball only goes up by 60% of what it did on the previous bounce. How high will the ball go after the 38th bounce? I took the 13.5x .6 for the 1st bounce which is 8.1Ft. and the 2nd bounce is 4.9ft. and the third is 2.9ft. I could do this for all the bounces but it would take me forever. How can I write a formula to just put in bounce to get height?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
I'll try to write a general formula.
After the 1st and succeeding bounces,
h+=+.6%2A13.5
h+=+.6%2A.6%2A13.5
h+=+.6%2A.6%2A.6%2A13.5
It looks like the formula would be
h+=+13.5%2A.6%5En where n is the number of bounces
and h is the height after that bounce
After 38 bounces,
h+=+13.5%2A.6%5E38
h+=+13.5%2A3.713%2A10%5E%28-9%29
h+=+5.0128%2A10%5E%28-8%29 ft
How much is that in millionths of an inch?
10%5E6%2A12+=+1.2%2A10%5E5 millionths of inch/ft
h+=+5.0128%2A10%28-8%29%2A1.2%2A10%5E5
h+=+6.015%2A10%5E%28-3%29
That looks like h+=+6.015 billionths
of an inch after the 38th bounce