Question 1021597
The infinite series summation is {{{a[1](1/(1-r))}}} where {{{a[1]}}} is the first term of the series, and r is the geometric multiplier. 

Say you were given 1+1/2+1/4+1/8+..... can you see that this adds to 2? Take 2 squares, the first is "1" of course then take the second and cut it in half with a line, then the remaining half to quarters. It reproduces the series, and from the equation -

{{{1(1/(1-.5))=2}}}

Now, your problem is tricky as there are 2 series to add, H(1+.8+.64+.512.....) this is vertical distance dropped. But you need to add the distance it goes up, H(.8+.64+.512.....) see how the first "up" is .8 the height of the drop? 

Let's solve the first series -

1+.8+.64+.512...    (we'll multiply by H after solving this)

{{{a[1]=1}}} and r=.8

{{{1(1/(1-.8))=5}}}

So the first series adds to 5H

The second series 

.8+.64+.512..

{{{a[1]=.8}}} and r=.8

{{{.8(1/(1-.8))=4}}} 

So this adds to 4H which makes sense, as the "up" is .8 times the prior "down".

Now, you've calculated that from a drop of H and a boucyness of .8, you get a total travel of 9H. 
One final step. They told you the travel was 63 meters.

9H = 63m

H=63/9 m

H= 7m 

Now you can solve an infinite number of infinite geometric series equations..... 


NOTE: The tutors here have no way to communicate with each other. I believe the other 2 answers are wrong.  Stanbon did not include the "up" distance. The problem states "total vertical distance traveled" .
Roth started at .8h which ignored the initial drop, the first full 1h. (Edit again - Roth has corrected and confirmed my answer)