Calculating Total Distance Traveled by an Elastic Ball with Infinite Bounces

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SUMMARY

The total distance traveled by an elastic ball dropped from a height of 2 meters, bouncing back to three-quarters of its previous height, is calculated using the concept of geometric series. The initial drop contributes 2 meters, while the subsequent bounces form a series with a first term of 3/2 meters and a common ratio of 3/4. The correct setup for the total distance is 2 + 2(S_n - 2), where S_n equals 8 meters, resulting in a total distance of 14 meters.

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Homework Statement


When dropped, an elastic ball bounces back up to a height three-quarters of that from which it fell. If the ball is dropped from a height 2 m and allowed to bounce up and down indefinitely, what is the total distance it travels before coming to rest?



Homework Equations



I think I have to use Partial sums of geometric series.

If r is not equal to 1 then


[tex]S_n = a + ar + ar^2 + ... ar^n-1 = a(1-r^n)/1-r[/tex]



The Attempt at a Solution



It's really easy to understand the question, but setting it up mathematecally is other story.

I tried to do 2 + 3/2 + 9/8 + 27/32 + 81/128 + ... + 3n/4n where a_1 = 2 and a_2 = 3/2
but this is not leading me to an final answer.

Trying to use the equation above saying a = 2 and r = 3/4

I get the final answer 8 m But the answer is 14 m.

What is the correct setup?
 
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If you use only 3/2+9/8+27/32+...+3n/4n where a_1=3/2, you get half the distance, save 2m [since it bounces up AND down the same distance every time except the first drop]
S_n=8 according to your calculations, so the new sum would be S_n-2

Thus you get the total distance as 2+2(S_n-2) = 2+12=14
 
duh?

Thank you for this eye opening reply.

Of course!
 

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