# ball bouncing forever

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• Aug 20th 2009, 08:22 AM
dat1611
ball bouncing forever
A ball drops from a height of 14 feet. Each time it hits the ground, it bounces up 35 percents of the height it fall. Assume it goes on forever, find the total distance it travels.

can anyone help me set this one up
• Aug 20th 2009, 08:51 AM
Twig
Hi

I aint particulary good at this, but IŽll give it a try.

First it drops 14 feet. Then it bounces 35 percent of the previous height with each bounce.

But it also drops that same distance as it rises with each bounce. So it actually moves two equal lengths each bounce.

So...maybe something like:

$14+\sum_{k=1}^{\infty}2\cdot 14\cdot \left(\frac{35}{100}\right)^{k}$

Which is a geometric sum.
• Aug 20th 2009, 08:54 AM
dat1611
ya thats right
• Aug 20th 2009, 09:02 AM
nikhil
here it is
let a be the height of fall (though a=14 feet)
on first jump height attained=.35a
now ball will fall through the same distance and after that it will raise to height=(.35)^2 (a) again it will fall through the same height and then it will raise to height=(.35)^3(a) this bouncing will happen infinite times so
the total distance will be given by
d=a+2a((.35)+(.35)^2+(.35)^3.............to infinite terms)
if we observe then terms in bracket forms a G.P whose sum to infinite terms is given by a/(1-r) where a is first term,r is common difference and|r|<1
so
d=14+28(.35/.65)
so
d=29.06 feet