A ball is dropped on the floor from a height 40 and it rebounds a distance .4h. What is the total distance the ball moves up and down.

The ball moves up by 0.4h every time it falls down, where
h is just previous  from which it falls.

So the distance
of  first fall  d1 = 40.

The 1st  rebound height d2=
40*(0.4)

Then d3 = 40*0*4. 

d4
= 0.40*(0.4)^2 and so forth.

So the total distance is
a series of up and down (fall and rebounds)  =
d1+d2+d3+d4.....

= 40+
(40*0.4+40.04)+(40*0.04^2+40*0.4^2)+....

=
40+2*40*(0*4+0.4^2+0.04^3+0.04^4+...)

=
40+240*0.4{1+0.4+0.4^2+0.4^3...)

= 40 +16(1-0.4)^(-1),  as
1+x+x^2+x^3 +x^4+... coverges to (1-x)^(-1) when x < 0. Here x =
0.4.

= 40+32/(0.6)

=
40+53.2

= 93.333.