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arroyjose

Conservation of energy

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I'm currently trying to solve this problem where a 85kg trampoline artist jumps vertically upward from the top of a platform with a speed of 5.5m/s ----- * <- artist | | 2 m high between the artist and the trampoline | ----- <- trampoline How fast is he going as he lands on the trampoline, 2m below? I'm using the formula for conservation of energy, but my algebra is giving me some problems. 1/2mv^2 + mgy1 = 1/2mv^2 + mgy2 m = mass v = speed/velocity g = gravity (9.8) y = height For my refernce frame I chose y1 to be 2 and y2 to be 0 I know that mass cancels so I end up with: 1/2(5.5)^2 + (9.8 * 2) = 1/2v^2 + 0 but I don't see I to derive it to solve for v Any help would be greatly aprreciated. [Edited by - arroyjose on October 17, 2004 3:25:38 PM]

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Yeap! My physics professor is way too smart and most of the time I have trouble following all the derivations he does to the formulas.

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Is this homework?

Anyway, why would the mass matter at all? As you know, a pea and a pound of lead will accelerate equally fast (modulo air friction).

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Obvious HomeworkTM

anyway.
You "somehow" got to
1/2(5.5)^2 + (9.8 * 2) = 1/2v^2
how you can get to such problems if you can't see that you need to multiply both sides by 2 to remove 1/2 from right ? so you get
(5.5)^2 + 9.8*4 = v^2
and
sqrt((5.5)^2 + 9.8*4)=v /// nitpicking: obviously, |v|
and if you want it to be nicely arranged,
v=sqrt((5.5)^2 + 9.8*4)
But if you can't do it yourself, i'd say, your education is completely screwed up and it woulda be alot better to do all math since 4th grade of elementary school (or maybe even 2nd).

edit: OMG,typoos[grin]

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Guest Anonymous Poster
2*2 = 4

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