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Quadratic Drag Question

  • Thread starter Oblio
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I know the fractions are technically the same thing as -1/g/sqrt[c/mg], but that doesnt help
 

learningphysics

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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]?
 
397
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In trying things out manually I'm finding that its sqrt[g] but i definitely didnt know that before now... hmm
 

learningphysics

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In trying things out manually I'm finding that its sqrt[g] but i definitely didnt know that before now... hmm
exactly... can you further simplify the integral?
 
397
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Its hard since its in the denominator in the square root...

is it close to...
sqrt[c/(msqrt[g])] lol thats wrong...
 

learningphysics

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Its hard since its in the denominator in the square root...

is it close to...
sqrt[c/(msqrt[g])] lol thats wrong...
[tex]\frac{-1}{g}\times\frac{1}{\sqrt{c/mg}}[/tex]

=

[tex]\frac{-1}{\sqrt{gc/m}}[/tex]

=

[tex]-\sqrt{\frac{m}{gc}}[/tex]
 
397
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I'll have to do some research on that, I can't say I'm following manipulating such distant numbers..
 
397
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[tex]\frac{-1}{g}\times\frac{1}{\sqrt{c/mg}}[/tex]

=

[tex]\frac{-1}{\sqrt{gc/m}}[/tex]

=

[tex]-\sqrt{\frac{m}{gc}}[/tex]
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

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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]
 
397
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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]

We often didn't use limits. Why this time?
 

learningphysics

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We often didn't use limits. Why this time?
Yeah, we don't need the limits... but we should add a constant when we take the integral...
 
397
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Ok so if arctan = that, I need to find out what the integral of arctan is..
 
397
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I realized that after I wrote it.

so subbing u back in I get...

-sqrt[m/gc]arctan [sqrt[c/mg] xv
 

learningphysics

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I realized that after I wrote it.

so subbing u back in I get...

-sqrt[m/gc]arctan [sqrt[c/mg] xv
yes...but you need a constant so:

[tex]t = -\sqrt{\frac{m}{gc}}arctan(\sqrt{\frac{c}{mg}}*v) + C[/tex]

we can solve for C by subsituting in t = 0, v = v0

so [tex]C = \sqrt{\frac{m}{gc}}arctan(\sqrt{\frac{c}{mg}}*v0)[/tex]

So [tex]t = -\sqrt{\frac{m}{gc}}arctan(\sqrt{\frac{c}{mg}}*v) + \sqrt{\frac{m}{gc}}arctan(\sqrt{\frac{c}{mg}}*v0)[/tex]
 
Last edited:
397
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Ok, and somehow my two square roots will boil down to v(ter)/g and vo/v(ter) ?
 

learningphysics

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EDIT:

actually no... you don't need to do this... sorry about that...

you can solve for vter... at the terminal velocity the net force is 0...

mg = cv^2

vter = sqrt(mg/c)

that should work...
 
Last edited:
397
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I only need the time to the top of the trajectory though?
Are you sure I need to find the time for down?
 

learningphysics

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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.
 
397
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yeah you're right. you don't need that part. I think you're almost done.

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]) ?
 

learningphysics

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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]) ?
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...
 
397
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Ok, one sec. But vter is when v=0 ?
 
397
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just C?
 

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