Confused about equation solving

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  • Thread starter JesseK
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  • #1
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Hey!

I'm confused about this:

½v21 = ½v20 - gh1

v21 = v20 - 2gh1

v1 = √v02 - 2gh1

Soo, what basically happened here when solving the equation for v1? I understand that the equation should be multiplied by ½ and then take a square root in order to solve for v1, but what's that factor "2" before gh (bold text). Where did that come from?

Thank you!
 

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  • #2
Ibix
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I understand that the equation should be multiplied by ½
No, you are multiplying by 2. Does that answer your question?
 
  • #3
berkeman
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As @Ibix says, multiply both sides of the equation by 2 (so the equal sign still holds -- always do the same operation on both sides of the equation when simplifying it):

[tex]2(\frac{1}{2}v_1^2) = 2(\frac{1}{2}v_0^2 - gh_1)[/tex]
 
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  • #4
Ibix
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1/2 * 1/2 = 1
What? If I have an apple and I cut it in half, and cut one if the halves in half, you expect the half of a half to be a whole apple?
 
  • #5
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What? If I have an apple and I cut it in half, and cut one if the halves in half, you expect the half of a half to be a whole apple?
Yeah, it was my bad, sorry. Obviously 1/2 * 1/2 = 1/4 haha :D
 
  • #6
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As @Ibix says, multiply both sides of the equation by 2 (so the equal sign still holds -- always do the same operation on both sides of the equation when simplifying it):

[tex]2(\frac{1}{2}v_1^2) = 2(\frac{1}{2}v_0^2 - gh_1)[/tex]
Oh, so every term on both sides of the equation should be multiplied by 2 (those brackets)? Now, I gotcha, thank youu! :D
 
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  • #7
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Hey!

I'm confused about this:

½v21 = ½v20 - gh1

v21 = v20 - 2gh1

v1 = √v02 - 2gh1
The line above confused me, at first, since it looks like you're taking the square root only of the ##v_0^2## term. You really should be using parentheses, like this: v1 = √(v02 - 2gh1)

Also, mathematically speaking, there will be two solutions for ##v_1##:
v1 = ±√(v02 - 2gh1)
JesseK said:
Soo, what basically happened here when solving the equation for v1? I understand that the equation should be multiplied by ½ and then take a square root in order to solve for v1, but what's that factor "2" before gh (bold text). Where did that come from?

Thank you!
 

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