# 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.

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### 2 Answers

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.

The distance travelled by the ball is 40 + 2*.4*40 +2*.4^2*40 +… Here we have a geometric progression as the height to which the ball rises after every bounce is a constant .4 which is less than 1. Also, the ball ideally will continue to bounce forever. So this becomes a determination of a GP with infinite terms.

The sum is equal to 40 + 2*.4*40 / (1-.4) = 40 + 80* .4 / .6 = 40 + 80* 4/6 = 40 + 160/3 = 280/3.

**The total distance traveled moving up and down is 280/3**