Originally posted by Drew LIt wont "travel" as it was dropped from height, if it was dropped perfectly vertical, it will rest where it first bounced. (in theory!)
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?
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for all you smart fellers
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Originally posted by smomofoyeah, it was on my pre-calculus final.
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.
Originally posted by smomofohahaha, I got that one right and pulled off an 83 on the exam
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.
Originally posted by smomofoThis is a correct approach, you just should show that the series converges.
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.