Physics Help Forum Manipulating Equations

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 Sep 3rd 2009, 06:20 PM #1 Junior Member     Join Date: Sep 2009 Posts: 19 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?
Sep 3rd 2009, 06:48 PM   #2
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Join Date: Apr 2008
Posts: 815
 Originally Posted by WhoCares357 "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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If the problem is too hard just let the Universe solve it.

Sep 3rd 2009, 07:03 PM   #3
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Posts: 19
 Originally Posted by arbolis 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 07:27 PM. Reason: Introducing latex

Sep 3rd 2009, 07:34 PM   #4
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Join Date: Apr 2008
Posts: 815
 Originally Posted by WhoCares357 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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Isaac
If the problem is too hard just let the Universe solve it.

 Sep 4th 2009, 10:58 AM #5 Banned   Join Date: Aug 2009 Location: UK Posts: 240 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
 Sep 4th 2009, 05:59 PM #6 Junior Member     Join Date: Sep 2009 Posts: 19 Okay thank you. I understand it now.
 Sep 5th 2009, 02:31 AM #7 Banned   Join Date: Aug 2009 Location: UK Posts: 240 Glad you understand, i tried to break it down so people could understand.

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