- 397
- 0
I know the fractions are technically the same thing as -1/g/sqrt[c/mg], but that doesnt help
what is [tex]\frac{g}{\sqrt{g}}[/tex]?I know the fractions are technically the same thing as -1/g/sqrt[c/mg], but that doesnt help
exactly... can you further simplify the integral?In trying things out manually I'm finding that its sqrt[g] but i definitely didnt know that before now... hmm
[tex]\frac{-1}{g}\times\frac{1}{\sqrt{c/mg}}[/tex]Its hard since its in the denominator in the square root...
is it close to...
sqrt[c/(msqrt[g])] lol thats wrong...
For the time being, I'm left with[tex]\frac{-1}{g}\times\frac{1}{\sqrt{c/mg}}[/tex]
=
[tex]\frac{-1}{\sqrt{gc/m}}[/tex]
=
[tex]-\sqrt{\frac{m}{gc}}[/tex]
yeah, use arctan. we should actually have limits on that integral... so: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
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]
Yeah, we don't need the limits... but we should add a constant when we take the integral...We often didn't use limits. Why this time?
That IS arctan isnt it?[tex]\int_{u_{initial}}^{u_{final}}[/tex] [tex]\frac{du}{1+u^2}[/tex]
yeah.That IS arctan isnt it?
no arctan IS the integral.Ok so if arctan = that, I need to find out what the integral of arctan is..
yes...but you need a constant so:I realized that after I wrote it.
so subbing u back in I get...
-sqrt[m/gc]arctan [sqrt[c/mg] xv
yeah you're right. you don't need that part. I think you're almost done.I only need the time to the top of the trajectory though?
Are you sure I need to find the time for down?
yeah you're right. you don't need that part. I think you're almost done.
yeah, look at post #92, with the constant evaluated... you want the time when v = 0. What time do you get? Try to sub in vter...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]) ?