Equation of Motion for a Projectile Under Quadratic Air Resistance

In summary: I rearrange this equation you get:v / [(c/mg)(v^3/3) + v^2] ?Yes. so will it be:v / [(c/mg)(v^3/3) + v^2] ?
  • #71
Besides moving it out of the integral?
 
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  • #72
Oblio said:
Besides moving it out of the integral?

yes. forget about moving it out of the integral...
 
  • #73
learningphysics said:
yes. forget about moving it out of the integral...

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...
 
  • #74
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...

can you do anything with the g that is there and the g inside the square root?

also try to clean up the fraction so that I don't have additional fractions inside the numerator and denominator... ie: in the numerator I want stuff being multiplied... in the denominator I want stuff being multiplied...
 
  • #75
I'm sure this isn't hard but I can't see anything...
 
  • #76
I know the fractions are technically the same thing as -1/g/sqrt[c/mg], but that doesn't help
 
  • #77
Oblio said:
I know the fractions are technically the same thing as -1/g/sqrt[c/mg], but that doesn't help

what is [tex]\frac{g}{\sqrt{g}}[/tex]?
 
  • #78
In trying things out manually I'm finding that its sqrt[g] but i definitely didnt know that before now... hmm
 
  • #79
Oblio said:
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?
 
  • #80
Its hard since its in the denominator in the square root...

is it close to...
sqrt[c/(msqrt[g])] lol that's wrong...
 
  • #81
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...

[tex]\frac{-1}{g}\times\frac{1}{\sqrt{c/mg}}[/tex]

=

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

=

[tex]-\sqrt{\frac{m}{gc}}[/tex]
 
  • #82
I'll have to do some research on that, I can't say I'm following manipulating such distant numbers..
 
  • #83
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]

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
 
  • #84
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

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]
 
  • #85
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]


We often didn't use limits. Why this time?
 
  • #86
Oblio said:
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...
 
  • #87
learningphysics said:
[tex]\int_{u_{initial}}^{u_{final}}[/tex] [tex]\frac{du}{1+u^2}[/tex]

That IS arctan isn't it?
 
  • #88
Oblio said:
That IS arctan isn't it?

yeah.
 
  • #89
Ok so if arctan = that, I need to find out what the integral of arctan is..
 
  • #90
Oblio said:
Ok so if arctan = that, I need to find out what the integral of arctan is..

no arctan IS the integral.
 
  • #91
I realized that after I wrote it.

so subbing u back in I get...

-sqrt[m/gc]arctan [sqrt[c/mg] xv
 
  • #92
Oblio said:
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:
  • #93
Ok, and somehow my two square roots will boil down to v(ter)/g and vo/v(ter) ?
 
  • #94
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:
  • #95
I only need the time to the top of the trajectory though?
Are you sure I need to find the time for down?
 
  • #96
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?

yeah you're right. you don't need that part. I think you're almost done.
 
  • #97
learningphysics said:
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]) ?
 
  • #98
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]) ?

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...
 
  • #99
Ok, one sec. But vter is when v=0 ?
 
  • #100
just C?
 
  • #101
Oblio said:
Ok, one sec. But vter is when v=0 ?

No. look at post #94 for vter.
 
  • #102
Oblio said:
Ok, one sec. But vter is when v=0 ?

v=0 is at the top of the trajectory.
 
  • #103
learningphysics said:
No. look at post #94 for vter.



Ok I thought you were saying that :P
 
  • #104
I got t=c for v=0
 
  • #105
and vter = sqrt[mg/c]
 

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