Question

A ball is dropped from a height of 21 m and always rebounds of the height of the previous drop. How far does it travel (up and down) before coming to rest?

Answer 1

you are going to have to sum a geometric series, but you are missing something in the question

Answer 2

what am i missing?

Answer 3

read your post

Answer 4

oh it rebounds 1/3 of the height of the previous drop

Answer 5

ok now we can do it
first off it drops 21 m, put that number aside, we will add in later

Answer 6

then it goes up 7 and also down 7 as \(21\times \frac{1}{3}=7\) so \[2\times
21\times \frac{1}{3}\] for the first bounce up and down

Answer 7

second bounce up and down will be \[2\times 21\times \left(\frac{1}{3}\right)^2\] or \[42\times \frac{1}{3^2}\]

Answer 8

and so on
add up \[42\times\frac{1}{3}+42\times \frac{1}{3^2}+42\times \frac{1}{3^2}+...\] or in sigma notation \[42\sum_{n=1}^{\infty}\frac{1}{3^n}\]