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Old Sep 3rd 2009, 05:20 PM   #1
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Manipulating Equations

"An object falls from a height h from rest. If it travels a fraction of the total height of 0.7212 in the last 1.00 s, find the time of its fall. Find the height of its fall."

A professor worked this for the class by using two different equations to create another one. I didn't understand it at all. I don't even know what some of his resulting variables represent.

Does any one know what I should do?
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Old Sep 3rd 2009, 05:48 PM   #2
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Originally Posted by WhoCares357 View Post
"An object falls from a height h from rest. If it travels a fraction of the total height of 0.7212 in the last 1.00 s, find the time of its fall. Find the height of its fall."

A professor worked this for the class by using two different equations to create another one. I didn't understand it at all. I don't even know what some of his resulting variables represent.

Does any one know what I should do?
Could you provide the work of your professor? So that we can maybe help you to understand what he did.
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Old Sep 3rd 2009, 06:03 PM   #3
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Originally Posted by arbolis View Post
Could you provide the work of your professor? So that we can maybe help you to understand what he did.
I don't have a scanner but I'll try to explain it as clearly as I can here:

The final formula is t'=t(1-sqrt(1-f)) where t' is the "last 1 second of the fall", t is the total time, f is h'/h (h'/h = .7212 in this problem) where h' is the distance traveled in the last 1 second and h is the total distance.

I understand most of his reasoning except for one step.
Here are his steps:
EQUATION ONE
1) $\displaystyle h=\frac{gt^2}{2}$
2) $\displaystyle t=\sqrt {\frac{2h}{g}}$
EQUATION TWO
3) $\displaystyle h-h'=\frac{g(t-t')^2}{2}$
4) This is where I'm confused. Somehow he used the two above equations to get the equation:
5) $\displaystyle (1-f)=\frac{(t-t')^2}{t^2}$
6) $\displaystyle (1-f)=(1-\frac{t'}{t})^2$
7) $\displaystyle 1-\frac{t'}{t}=\sqrt {1-f}$
8) $\displaystyle t'=t(1- \sqrt {1-f})$

By the way, I figured out how to plug in the correct constants into his equation to get the correct time. I used the time to find the distance. However, I would still like to know how he derived his equation.

Last edited by arbolis; Sep 3rd 2009 at 06:27 PM. Reason: Introducing latex
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Old Sep 3rd 2009, 06:34 PM   #4
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Originally Posted by WhoCares357 View Post
I don't have a scanner but I'll try to explain it as clearly as I can here:

The final formula is t'=t(1-sqrt(1-f)) where t' is the "last 1 second of the fall", t is the total time, f is h'/h (h'/h = .7212 in this problem) where h' is the distance traveled in the last 1 second and h is the total distance.

I understand most of his reasoning except for one step.
Here are his steps:
EQUATION ONE
1) $\displaystyle h=\frac{gt^2}{2}$
2) $\displaystyle t=\sqrt {\frac{2h}{g}}$
EQUATION TWO
3) $\displaystyle h-h'=\frac{g(t-t')^2}{2}$
4) This is where I'm confused. Somehow he used the two above equations to get the equation:
5) $\displaystyle (1-f)=\frac{(t-t')^2}{t^2}$
6) $\displaystyle (1-f)=(1-\frac{t'}{t})^2$
7) $\displaystyle 1-\frac{t'}{t}=\sqrt {1-f}$
8) $\displaystyle t'=t(1- \sqrt {1-f})$

By the way, I figured out how to plug in the correct constants into his equation to get the correct time. I used the time to find the distance. However, I would still like to know how he derived his equation.
Ah I see. From 4) to 5) he divided by $\displaystyle h$ on the left side and by $\displaystyle \frac{gt^2}{2}$ on the right hand side, which is equal to h.
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Old Sep 4th 2009, 09:58 AM   #5
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h = (1/2)at^2

t = t1 / (1-(1-(Sq-rt(1-0.7212)) = 1/(1-(sq-rt(1-0.7212)) = 2.1187 s

h= 21.9958 m

h1 = h-h2

h2= t2 = t-t1 = 2.1187-1 = 1.1187 s

h2= (1/2)a(t2)^2 = 6.1323 m

h1= 21.9958-6.1323 = 15.8635 m

f = h1/h = 0.7212
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Old Sep 4th 2009, 04:59 PM   #6
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Okay thank you. I understand it now.
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Old Sep 5th 2009, 01:31 AM   #7
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Glad you understand, i tried to break it down so people could understand.
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