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NateRiver
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20 Jan 2013, 12:22 pm

v^2= w^2(a^2-x^2)

e= mgh+ 1/2mv^2



TallyMan
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20 Jan 2013, 12:26 pm

NateRiver wrote:
v^2= w^2(a^2-x^2)

e= mgh+ 1/2mv^2


v^2 / w^2 = a^2 - x^2

v^2 / w^2 - a^2 = - x^2

x^2 = - v^2 / w^2 + a^2

x = square root(- v^2 /w^2 + a^2)


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Last edited by TallyMan on 20 Jan 2013, 12:33 pm, edited 2 times in total.

ruveyn
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20 Jan 2013, 12:26 pm

NateRiver wrote:
v^2= w^2(a^2-x^2)

e= mgh+ 1/2mv^2


Solve for v in the first equations and plug it into the second. All that is left is x a variable and g a constant.

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thomas81
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20 Jan 2013, 12:27 pm

NateRiver wrote:
v^2= w^2(a^2-x^2)


a^2-x^2 =(v^2/w^2)

x^2 =a^2-(v^2/w^2)

x=sqrt(a^2-(v^2/w^2))
NateRiver wrote:
e= mgh+ 1/2mv^2


mgh = e - 1/2mv^2

g = (e -1/2mv^2)/mh



I think


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Trencher93
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20 Jan 2013, 1:09 pm

I would be more interested in why you want to do this. Why are you asking? Just dumping equations into a post looks like you're asking other people to do your homework for you. I would encourage people not to help with that sort of thing, since it defeats the purpose of learning. If someone can't solve simultaneous equations like this, the only way to learn is to do it step-by-step. If you're genuinely lost, asking "why" questions would be more helpful than just having people type in solutions that book examples probably already show, since you're probably lost in reading the book examples.

If you don't know that g is a constant, you're definitely just beginning to learn whatever it is you're trying to learn.

https://en.wikipedia.org/wiki/Gravitational_constant

Either that or you definitely need to provide more information.



thomas81
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20 Jan 2013, 4:25 pm

Trencher93 wrote:

If you don't know that g is a constant, you're definitely just beginning to learn whatever it is you're trying to learn.

https://en.wikipedia.org/wiki/Gravitational_constant

Either that or you definitely need to provide more information.

isnt that context specific?

Isnt it that g is a constant only in applied mathematics while in pure mathematics the meaning of g is arbitrary?


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Trencher93
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20 Jan 2013, 6:22 pm

thomas81 wrote:
isnt that context specific?


Sure, but e= mgh+ 1/2mv^2 is definitely context specific. It's a potential energy equation from physics.

http://www.real-world-physics-problems. ... nergy.html

That's what I'm saying - this is some sort of homework assignment, and better than just throwing out answers, we should help the student understand the material, but more effort needs to be made by the student than just posting an equation without explanation.

In this case, g is a constant. Apparently the student wants to solve the equation for a constant, and you can't do that in math. Many of these "answers" are misleading for that reason, since g is not a variable. ruveyn's answer (and similar ones) are correct.