The Algebra Symposium: Notes on Bouncing Balls
This file is http://www2.math.uic.edu/~jlewis/as/asbouncenote.pdf
From Calculus, Hughes–Hallett, et al.:
3
1. A ball is dropped from a height of 10 feet and bounces. Each bounce is of the height
4
of the bounce before.
Thus after the ball hits the floor for the first time, the ball rises to
3
= 7.5 feet, and after the it hits the floor for the second time, the ball
a height of 10
4
2
3
3
= 10
= 5.625 feet.
rises to a height of 7.5
4
4
(a) Find an expression for the height to which the ball rises after it hits the floor for the nth
time.
n
3
th
After it hits the floor for the n time, the ball rises to a height of 10
feet.
4
(b) Find an expression for the total vertical distance the ball has traveled when it hits the
floor for the first, second, third, and fourth times.
After
10 + 2 ·
one
1 time the ball has traveled 10 feet. After two times theball
1has traveled
2
3
3
3
10
feet. After three times the ball has traveled 10 + 2 · 10
+ 2 · 10
feet.
4
4
4
(c) Find an expression for the total vertical distance the ball has traveled when it hits the
floor for the nth time. Express your answer in a closed form.
Hint
1 + r + r2 + . . . + rn =
1 − rn+1
.
1−r
n
3
to the total
4
vertical distance. Thus the total vertical distance traveled after hitting for the nth time is
1
n−1 !
3
3
10 1 + 2 ·
+ ... + 2 ·
.
4
4
After hitting the floor for the nth time, the next bounce adds 2 · 10
A closed form of this expression is
10 ·
1+2
n−1
X
k=1
3
4
k !
n
(3/4) − (3/4)
= 10 1 + 2
1 − (3/4)
asbouncenote.tex 1/2
If we let n → ∞, the total vertical distance traveled is
(3/4)
10 1 + 2
= 70feet.
1 − (3/4)
2. You might think that the ball [in the previous problem] keeps bouncing forever since it
takes infinitely many bounces.
Is this true?
Although there are infinitely many bounces, they occur in finite time. If a ball falls from
a height h feet, its vertical position at time t is given by
1
y = h − gt2 ,
2
2
where g ≈ 32 feet/(sec) is the gravitational constant. Thus y = 0 when
p
t = 2h/g
p
√
= 2/g h
1√
≈
h.
4
An argument similar to the calculation of the total total vertical distance traveled will
show that the total time elapsed traveled after hitting for the nth time is
r p
p
1
n−1 2
10
1+2·
3/4 + . . . + 2 ·
3/4
.
g
A closed form of this expression is
!
r
r
n−1
k
X p
2
2
10
· 1+2
3/4
= 10
g
g
p n!
p
3/4 − 3/4
p
1+2
1 − 3/4
k=1
If we let n → ∞, the total time elapsed is
!
p
r
3/4
2
p
10
1+2
≈ ∗sec.
g
1 − 3/4
We exploit the simultaneous convergence of the two geometric series:
∞
X
k
(bounce ratio) ,
k=1
∞ √
X
bounce ratio
k
.
k=1
asbouncenote.tex 2/2