A ball is dropped from a height of 10 feet and bounces. Each
bounce is [3/4] of the height of the bounce before. Thus
after the ball hits the floor for the first time, the ball rises
to a height of 10([3/4]) = 7.5 feet, and after the it
hits the floor for the second time, the ball rises to a height of
7.5([3/4]) = 10([3/4])2 = 5.625 feet.
(a)
Find an expression for the height to which the ball
rises after it hits the floor for the nth time.
After it hits the floor for the nth time, the ball
rises to a height of 10([3/4])n feet.
(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 one time the ball has traveled 10 feet. After two times
the ball has traveled 10 + 2·10([3/4])1
feet. After three times the ball has traveled 10 + 2·10([3/4])1 + 2·10([3/4])2
feet.
(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
.
After hitting the floor for the nth time, the next
bounce adds 2·10([3/4])n to the total
vertical distance. Thus the total vertical distance traveled after
hitting for the nth time is
10
⎛ ⎝
1 + 2·
⎛ ⎝
3
4
⎞ ⎠
1
+ …+ 2·
⎛ ⎝
3
4
⎞ ⎠
n−1
⎞ ⎠
.
A closed form of this expression is
10 ·
⎛ ⎝
1 + 2
n−1 ∑ k=1
⎛ ⎝
3
4
⎞ ⎠
k
⎞ ⎠
= 10
⎛ ⎝
1 + 2
(3/4)− (3/4)n
1 − (3/4)
⎞ ⎠
If we let n → ∞, the total vertical distance traveled is
10
⎛ ⎝
1 + 2
(3/4)
1 − (3/4)
⎞ ⎠
= 70 feet.
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
y
= h −
1
2
g t2,
where g ≈ 32 feet/(sec)2 is the
gravitational constant. Thus y = 0 when
t
=
√
2h/g
=
√
2/g
√h
≈
1
4
√h.
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
10
⎛ √
2
g
⎛ ⎝
1 + 2·
⎛ ⎝
√
3/4
⎞ ⎠
1
+ …+ 2·
⎛ ⎝
√
3/4
⎞ ⎠
n−1
⎞ ⎠
.
A closed form of this expression is
10
⎛ √
2
g
·
⎛ ⎝
1 + 2
n−1 ∑ k=1
⎛ ⎝
√
3/4
⎞ ⎠
k
⎞ ⎠
= 10
⎛ √
2
g
⎛ ⎝
1 + 2
√
[ˉ(3/4)]− √{3/4}n
1 −
√
[ˉ(3/4)]
⎞ ⎠
If we let n → ∞, the total time elapsed is
10
⎛ √
2
g
⎛ ⎜
⎝
1 + 2
√
3/4
1 −
√
3/4
⎞ ⎟
⎠
≈ * sec.
We exploit the simultaneous convergence of the two geometric
series:
∞ ∑ k=1
(bounceratio)k,
∞ ∑ k=1
⎛ ⎝
√
bounceratio
⎞ ⎠
k
.
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