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## Wednesday, 27 March 2013

### The Geometric Series Application: A Bouncing Ball

Suppose you drop a basketball from a height of 10 feet. After it hits the floor, it reaches a height of 7.5 = 10 .  feet; after it his the floor for the second time, it reaches a height of 5.625 = 7.5 .  = 10 .  feet, and so on and so on.
Does the ball ever come to rest, and if so, what total vertical distance will it have traveled?

This will lead to summing a geometric series, but let us first investigate, what happens to a ball being dropped from a height h.
Since the ball is subject to free fall, at time t (in seconds) the ball will be at height

h(t) = h - t2

feet above ground, as long as h(t 0. Consequently the time t0 it takes the ball to hit the floor is given by
h = t02,

and so it follows that
t0 = .

At the surface of the earth, g  32 ft/sec2.Let dn be the distance (in feet) the ball has traveled when it hits the floor for the nth time, and let tn be the time (in seconds) it takes the ball to hit the floor for the nth time.
Clearly d1 = 10. After the ball has hit the floor for the first time it rises 10 .  feet and then drops the same distance. Consequently

d2 = 10 + 2 . 10 . .

After the ball has hit the floor for the second time it rises 10 .  feet and then drops the same distance. Consequently
d3 = 10 + 2 . 10 .  + 2 . 10 . .

In general, when the ball hits the floor for the nth time it will have traveled a total vertical distance of
dn = 10 + 2 . 10 .  + 2 . 10 .  + ... + 2 . 10 . .

The total distance d traveled by the ball is therefore given by the geometric series
d = 10 + 2 . 10 .  + 2 . 10 .  + ...

We can rewrite this as
d = - 10 + 2 . 101 +  +  + ...  = - 10 + 20 = 70.

The ball travels a total vertical distance of 70 feet!Let's do the same kind of computations for time: We already computed that

t1 =  = .

After the ball has hit the floor for the first time it rises 10 .  feet and then drops the same distance. Note that it is easy to see that it takes the same time for the ball to rise to its maximum height as it takes the ball to return to the ground. Consequently
t2 =  + 2 . .

I leave it to you to check the details of the general formula
t =  + 2 .  + 2 .  + ...

Once again this is a geometric series; rewriting yields
t =  + 2 .   +  + ...

Thus
t =  + 2 .  .   11.0112

seconds.Read this again slowly: Even though the ball bounces infinitely often, it comes to rest after a little more than 11 seconds!
Below you find a table for dn and tn for t = 1,..., 50:

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50