- 15 Jun '07 06:01

It wont "travel" as it was dropped from height, if it was dropped perfectly vertical, it will rest where it first bounced. (in theory!)*Originally posted by Drew L***A rubber ball dropped 80 feet rebounds on each bounce 3/5 of the height from which it fell. How far will it ravel before coming to rest?**

for all you smart fellers

- 15 Jun '07 06:34

yeah, it was on my pre-calculus final.*Originally posted by smomofo***Cool. I actually figured that out on my own! Thanks.**

A1 / 1 - r where A1 = initial drop and r = rate of rebound

First find the height of the first rebound

80 * (3/5) = 48

Second Use the geometric infinite geometric convergent formula [ once for the drop and once for the rebound, drop is given, step one finds the rebound]

80 / 1 -(3/5) = 80 / (2/5) = 80 * (5/2) = 200

48 / 1 -(3/5) = 48 / (2/5) = 48 * (5/2) = 120

Finally, add the two numbers together and there you have it. - 15 Jun '07 06:41

hahaha, I got that one right and pulled off an 83 on the exam*Originally posted by smomofo***I couldn't remember that stuff, so I just added the first few terms of (0.6) + (0.6)^2 + (0.6)^3 + ... and saw that it was headed for 1.5 and went from there. Maybe not the way it's supposed to be done, but I bet I could have faked my way to a decent mark on that question.** - 15 Jun '07 06:58

This is a correct approach, you just should show that the series converges.*Originally posted by smomofo***I couldn't remember that stuff, so I just added the first few terms of (0.6) + (0.6)^2 + (0.6)^3 + ... and saw that it was headed for 1.5 and went from there. Maybe not the way it's supposed to be done, but I bet I could have faked my way to a decent mark on that question.**