Sequences/Series

JQ2009 · #1 $Old$ October 18th, 2009, 11:58 AM

I'm not sure if this is the right forum for this so I appologise if this is the wrong place.

I think I have worked out the answer to my problem but I'm not sure I have done it right. My question is:

Someone bungee jumping falls 120 feet before being pulled back up by the cord. They always rebound 1/3 of the distance they have fallen and then fall 2/3 of the distance of the last rebound. How far do they travel?

I got 204.77 feet but I am unsure I have done it correctly.

Grandad · #2 $Old$ October 19th, 2009, 12:08 AM

Hello JQ2009

Quote:

Originally Posted by JQ2009 $View Post$

I'm not sure if this is the right forum for this so I appologise if this is the wrong place.

I think I have worked out the answer to my problem but I'm not sure I have done it right. My question is:

Someone bungee jumping falls 120 feet before being pulled back up by the cord. They always rebound 1/3 of the distance they have fallen and then fall 2/3 of the distance of the last rebound. How far do they travel?

I got 204.77 feet but I am unsure I have done it correctly.

The total distance is the sum of two infinite GP's (which we can then combine into one), one for the downward movement and one for the upward movement, as follows:

The first downward movement covers a distance of 120 feet, and the first rebound is $\tfrac13$ of this, namely 40 feet. Each subsequent downward movement is then $\tfrac23\cdot\tfrac13=\tfrac29$ of the previous one, and each upward movement (rebound) is $\tfrac13\cdot\tfrac23=\tfrac29$ of the previous rebound. So the total distance travelled is:

$S = 120 + \tfrac29\cdot120+(\tfrac29)^2\cdot120+...40 + \tfrac29\cdot40+(\tfrac29)^2\cdot40+...$

$=160 + \tfrac29\cdot160 +(\tfrac29)^2\cdot 160 + ...$

which is then an infinite GP with $a = 160, r=\tfrac29$ .

So $S = \frac{a}{1-r}=\frac{160}{1-\tfrac29}=\frac{160*9}{7}=205.7$ , which is close to your answer, but not quite the same!

Grandad