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 
t^{2}
feet above
ground, as long as
h(t) 0. Consequently the time t_{0} it
takes the ball to hit the floor is given by
h =
t_{0}^{2},
and so it follows that
t_{0} =
.
At the surface of the earth,
g 32 ft/sec^{2}.
Let d_{n} be the distance (in feet) the ball has traveled when it
hits the floor for the nth time, and let t_{n} be the time (in
seconds) it takes the ball to hit the floor for the nth time.
Clearly d_{1} = 10. After the ball has hit the floor for the first
time it rises
10^{ . } feet and then drops the same
distance. Consequently
d_{2} = 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
d_{3} = 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
d_{n} = 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
^{ . }10
1 +
+
+
^{ ... } =  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
t_{1} =
=
.
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
t_{2} =
+ 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 d_{n} and t_{n} for
t = 1,..., 50:



1



2



3



4



5



6



7



8



9



10



11



12



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17



18



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20



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25



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27



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31



32



33



34



35



36



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38



39



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50



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