A superball dropped from a height of 40ft always rebounds 1/5 of the distance fallen. approximately how far does the ball travel before coming to rest? Please help :(
geometric series for this one leave the initial drop of 40 out of it, we can put that in later
drops 40 put that aside bounces up \(40\times \frac{1}{5}\) and down \(40\times \frac{1}{5}\) for a total of \[80\times \frac{1}{5}\]
next one is \[80\times( \frac{1}{5})^2\]
and so on you have to add \[80\times \frac{1}{5}+80\times (\frac{1}{5})^2+80\times (\frac{1}{5})^2+...\]
Do you know the final answer?
i can compute it, yes does this summing a geometric series look familiar?
yes it's the section I'm on now
we can make life a little easier if we make it \[20+20\times \frac{1}{5}+20\times (\frac{1}{5})^2+...\]
or even easier as \[20(1+\frac{1}{5}+(\frac{1}{5})^2+(\frac{1}{5})^2+...\] formula for the second part is \[\frac{1}{1-r}\] where \(r=\frac{1}{5}\)
you get \[\frac{1}{1-\frac{1}{5}}=\frac{1}{\frac{4}{5}}=\frac{5}{4}\] now multiply that by \(20\) then add that \(40\) for the first drop
\[20\times \frac{5}{4}=25\]if my arithmetic is correct, making your answer \(65\)
Thank you for your help, but unfortunately that was incorrect
really? maybe i made a mistake we can try again if you like
lets go slow it goes up \(40\times \frac{1}{5}\) and down \(40\times \frac{1}{5}\) for a total of \[8+8=16\] oh damn
i said 20 and it was 16 because i cannot divide properly lets try \[16\times \frac{5}{4}=20\]
for a total of \(60\)
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