Expert: Scott A Wilson - 1/22/2012

Question
Hello, I need help with this extra credit problem below:

A ball is dropped from a height of 10m. It rebounds 1/2 the distance on each bounce. What is the total diastance the ball travels?(Round your anwser the nearest whole number)

Thanks for your help. I really need the anwser soon so please anwser quickly!:D Please, and thanks again.

Answer
It can be seen that the 1st fall is 10m, since that is the distance to the ground.

It can be seen that since the ball goes half as tall, it goes up to 5m on the 2nd bounce.
Once up there, it drops the 5m again.  This makes it go a total of 15m down plus
5m up.

The third bounce it will have risen to a height of 2.5.
That makes the distance down be 10+5+2.5 and the distance up be 5+2.5.

As this is repeated, we get the distance down to be 10S and the distance up to be 5S,
where S = sum(n=0 to n=infinity)(1/2^n).

Next please note that the value of 2S = 2*sum(n=0 to n=infinity)(1/2^n).

The 2 can be taken inside, giving 2S = sum(n=0 to n=infinity)(2/2^n).

Cancelling 2's in each term gives 2S = sum(n=0 to n=infinity)(1/2^(n-1)).

Taking out the term at n = 0 gives 2S = 1/2^-1 + sum(n=1 to n=infinity)(1/2^(n-1)).

Noting that 1/2^-1 = 2 and letting m=n-1 gives 2S = 2 + sum(m=0 to m=infinity)(1/2^m).

Next, note that the value of sum(m=0 to m=infinity)(1/2^m) is no different than
sum(n=0 to n=infinity)(1/2^n), and that is S.

This means we have 2S = 2 + S.  Subtracting S from both sides gives S = 2.

Since the total distance down was 10S and the total distance up was 5S,
that gives a total of 15S.  Since S = 2, 15S = 30.