Bouncy problem from Prentice Hall Chemistry textbook
adsims2001
edition):
(page 35, problem #80) A certain ball when dropped from any height,
bounces one-half the original height. If the ball was dropped from a
height of 60 in. and allowed to bounce freely, what is the total
distance the ball has traveled when it hits the ground for the third
time? Assume the ball bounces straight up and down.
So, I came up with this equation to (sort of) automate the process:
n = p + 1.5o(2/2^b)
(n = next distance, p = previous distance, o = original height, b =
number of bounces performed)
You have to repeat the arithmetic for every bounce...
First bounce (the exponential variable b is replaced by 1 for first)
n = 0 + 1.5(60)(2/2^1)
n = 90
Second bounce
n = 90 + 1.5(60)(2/2^2)
n = 135
Third bounce
n = 135 + 1.5(60)(2/2^3)
n = 157.5
This method gets the job done, but it takes a lot of paper and time.
I have been thinking hard on making a single equation, something like
this:
n = [algebra magic] + 1.5o(2/2^b)
I'm just a sophomore in American high school, so I was wondering if
someone could help me out? Thanks for any thinking you may put into
this.
Darrell
news:1188543506....@i13g2000prf.googlegroups.com...
There is a flaw in your reasoning. Look at it step by step.
Initial drop: 60 in down.
bounce1: 30 in up
30 in down
bounce2: 15 in up
15 in down. This is the third time it hits the ground.
150" total distance travelled.
For a general formula, consider you start at h then go h/2, h/2, h/4, h/4.
Find a pattern.
--
Darrell
Frederick Williams
>
> I came upon this problem in my textbook (Prentice Hall Chemistry 2005
> edition):
>
> (page 35, problem #80) A certain ball when dropped from any height,
> bounces one-half the original height. If the ball was dropped from a
> height of 60 in. and allowed to bounce freely, what is the total
> distance the ball has traveled when it hits the ground for the third
> time? Assume the ball bounces straight up and down.
60 (down) + 30 (up) + 30 (down) + 15 (up) + 15 (down) inches
= 150 inches.
No algebra, just arithmetic.
>
> So, I came up with this equation to (sort of) automate the process:
>
> n = p + 1.5o(2/2^b)
>
> (n = next distance, p = previous distance, o = original height, b =
> number of bounces performed)
>
> You have to repeat the arithmetic for every bounce...
>
> First bounce (the exponential variable b is replaced by 1 for first)
> n = 0 + 1.5(60)(2/2^1)
> n = 90
>
> Second bounce
> n = 90 + 1.5(60)(2/2^2)
> n = 135
>
> Third bounce
> n = 135 + 1.5(60)(2/2^3)
> n = 157.5
>
> This method gets the job done, but it takes a lot of paper and time.
> I have been thinking hard on making a single equation, something like
> this:
> n = [algebra magic] + 1.5o(2/2^b)
>
> I'm just a sophomore in American high school, so I was wondering if
> someone could help me out? Thanks for any thinking you may put into
> this.
--
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"He that giveth to the poor lendeth to the Lord, and shall be repaid,"
said Mrs Fairchild, hastily slipping a shilling into the poor woman's
hand.
adsims2001
Thanks all, we went over this in class and no one had the right
answer, heh. I do see what I did incorrectly now. Maybe I'll come
back to it later and try to finally make an equation that works...
Barb Knox
adsims2001 <adsim...@gmail.com> wrote:
Prove that 1 + a^1 + a^2 + ... + a^n = (a^(n+1) - 1)/(a - 1)
Then use that formula for the problem of calculating the distance for N
bounces.
BTW, what does this problem conceivably have to do with chemistry??
--
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Brian VanPelt
Here's an interesting side note to the problem. Suppose the ball
never stops bouncing, but always reaches a height half of its previous
bounce.
Forget about the 60 inches for now, suppose, for convenience, it
initially bounced 1 inch. So, the total distance traveled would be
2 inches for the initial bounce (count both up and down)
1 inch for the next full bounce
1/2 inch for the next bounce after that
and so on.
Add this all up.
2 + 1 + 1/2 + 1/4 + ... forever.
Barb showed you the way...
1 + 1/2 + 1/4 + ... (1/2)^n = (1 - (1/2)^(n + 1))/( 1/2 - 1)
Sorry Barb, I am stuck with doing things in my order because I am
stupid.
Anyways, look next at
(1 - x^(n + 1))/(1 - x)
where -1 < x < 1. Isn't it true that x^(n+1) diminishes to zero as n
gets bigger and bigger? Thus, the quantity
(1 - x^(n + 1))/(1 - x)
would have to get closer and closer to
1/(1 - x)
as n gets bigger and bigger.
So, you multiply by 2 (or 60, whichever is convenient).
Brian
Insomnia has a way of making people not sleep.
adsims2001
> In article <1188592276.873288.285...@22g2000hsm.googlegroups.com>,
The problem was an example of how to apply objective problem solving,
which we are going over before we start any actual chemistry.
Barb Knox
adsims2001 <adsim...@gmail.com> wrote:
> The problem was an example of how to apply objective problem solving,
> which we are going over before we start any actual chemistry.
I know you meant that to be a helpful answer, but I really do not
understand. Firstly, is "objective" problem solving something special,
as for example contrasted with other forms of problem solving
("subjective" maybe)? Secondly, haven't you already been exposed to
scads of such word-problems in mathematics classes?
Frederick Williams
In truth it's an exercise in adding a few numbers, something they
learned to do in primary school, but how do you get students to pay for
that?
Barry Schwarz
Perhaps the half-life of radioactive isotopes and total amount of
radiation put out over time?
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