How Many Bounces Until the Ball Reaches 2.44m?

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A ball dropped from a 6.10m building loses 10% of its kinetic energy with each bounce and needs to reach a height of 2.44m. The discussion explores using gravitational potential energy and kinetic energy equations to determine the number of bounces before the ball can no longer reach the required height. A geometric series approach is suggested, with the common ratio being 0.9 due to energy loss. While trial and error is mentioned as a method to find the solution, participants agree that calculating bounce heights one at a time may be simpler. Ultimately, it is estimated that the ball can bounce around ten times before failing to reach 2.44m.
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Homework Statement



A ball is dropped from rest from the top of a 6.10m building, falls straight down, collides inelastically with the ground and bounces back. The ball loses 10% of it's K.E every time it hits the ground. How many bounces can happen and the ball still reach a height of 2.44m above the ground.

Homework Equations



mgh = 1/2 mv2

xn = arn-1 (finding a term in a geometric series)

The Attempt at a Solution



I can come to an answer easily enough using the two equations stated above and the ideas of gravitational P.E being converted to K.E (so mass cancels out) and constructing a geometric series that uses r = 0.9 to accommodate the energy loss.

I have to essentially just guess terms until I'm in the right region and then increase or decrease my term until I reach the right answer. This seems somewhat messy to me, using trial and error.

Is there a cleaner method, that will use the information of the final height I need it to reach?

I'm just curious,

thanks!
 
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##x_n=ar^{n-1}##
##r^{n-1}=x_n/a##
##(n-1)logr=log x_n-loga##
 
Enigman said:
##x_n=ar^{n-1}##
##r^{n-1}=x_n/a##
##(n-1)logr=log x_n-loga##

Ahh, thank you!
 
x_n=x_0r^n But, it's much easier just to do 1 bounce at a time. It couldn't take more than about 10 bounces with r = .9. Or, do it using: every two bounces is 0.81, every 3 bounces is 0.729.
 
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