- #71
Oblio
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- 0
Besides moving it out of the integral?
Oblio said:Besides moving it out of the integral?
learningphysics said:yes. forget about moving it out of the integral...
Oblio said:we have 2 fractions multiplied together...
[tex]\frac{-1}{g}[/tex] x [tex]\frac{1}{/sqrt{c/mg}}[/tex]
I would think the sqrt would need to removed maybe...
Oblio said:I know the fractions are technically the same thing as -1/g/sqrt[c/mg], but that doesn't help
Oblio said:In trying things out manually I'm finding that its sqrt[g] but i definitely didnt know that before now... hmm
Oblio said:Its hard since its in the denominator in the square root...
is it close to...
sqrt[c/(msqrt[g])] lol that's wrong...
learningphysics said:[tex]\frac{-1}{g}\times\frac{1}{\sqrt{c/mg}}[/tex]
=
[tex]\frac{-1}{\sqrt{gc/m}}[/tex]
=
[tex]-\sqrt{\frac{m}{gc}}[/tex]
Oblio said:For the time being, I'm left with
[tex]-\sqrt{\frac{m}{gc}}[/tex] [tex]\int[/tex] [tex]\frac{du}{1+u^2}[/tex]
Now integrate it I assume
learningphysics said:yeah, use arctan. we should actually have limits on that integral... so:
[tex]-\sqrt{\frac{m}{gc}}[/tex] [tex]\int_{u_{initial}}^{u_{final}}[/tex] [tex]\frac{du}{1+u^2}[/tex]
Oblio said:We often didn't use limits. Why this time?
learningphysics said:[tex]\int_{u_{initial}}^{u_{final}}[/tex] [tex]\frac{du}{1+u^2}[/tex]
Oblio said:That IS arctan isn't it?
Oblio said:Ok so if arctan = that, I need to find out what the integral of arctan is..
Oblio said:I realized that after I wrote it.
so subbing u back in I get...
-sqrt[m/gc]arctan [sqrt[c/mg] xv
Oblio said:I only need the time to the top of the trajectory though?
Are you sure I need to find the time for down?
learningphysics said:yeah you're right. you don't need that part. I think you're almost done.
Oblio said:lol phew I was worried for a sec.
have I already defined the relation that will give me
t(top) = [tex]\frac{v(ter)}{g}[/tex]arctan([tex]\frac{vo}{v(ter)}[/tex]) ?
Oblio said:Ok, one sec. But vter is when v=0 ?
Oblio said:Ok, one sec. But vter is when v=0 ?
learningphysics said:No. look at post #94 for vter.