# Falling Bodies with air resitance

1. Jan 11, 2005

### BigStelly

I know this may be seen as a physics question and it is, but i am just looking for an explicit solution to falling bodies with air resistance.

OK many of you are probably familiar with the equation for a body falling with air resistance.

mdv/dt=mg-kv^2
I am not sure exactly how to solve for v in terms of t, It seems to be seperable but i have problems with the integration.
mdv=(mg-kv^2)dt

If an explicit solution is indeed possible then i would like to know because I am unsure of my results on this one. Thanks!

2. Jan 11, 2005

### Dr Transport

$$\frac{dv}{mg-kv^{2}} = \frac{dt}{m}$$, now integrate this function and apply the boundary conditions.

3. Jan 11, 2005

### HallsofIvy

Hint: mg- kv2 can be factored as $(\sqrt{mg}-\sqrt{k}v)(\sqrt{mg}+\sqrt{k}v)$. You can use "partial fractions" to integrate it.

Last edited by a moderator: Jan 12, 2005
4. Jan 11, 2005

### BigStelly

Sorry guys i should have clarified the integration is what is getting me, help would be greatly appreciated.....

5. Jan 12, 2005

### marlon

$$\frac{1}{mg-kv^2} = \frac{A}{\sqrt{mg} - \sqrt{k}v} + \frac{B}{\sqrt{mg} + \sqrt{k}v}$$

Determin A and B and then integrate these two fractions...

marlon

6. Jan 12, 2005

### dextercioby

It would have been nicer,if we had some Stokes force there.It would have given it a little spice... :tongue2:

Daniel.

7. Jan 12, 2005

### Dr Transport

the integral is $$\int \frac{dv}{\sqrt{mg - kv^{2}}} = \frac{1}{\sqrt{k}}\arcsin\left(\sqrt{\frac{ k}{mg}}v\right)$$

There is no need to break the fraction into partial fractions to integrate.

8. Jan 12, 2005

### vincentchan

Drtransport.... where did the squareroot come from

9. Jan 12, 2005

### dextercioby

Nope,wrong formula i'm afraid.no square root over the denominator,no harmonic motion whatsoever...

Daniel.

10. Jan 13, 2005

### Dr Transport

OOOPPPS,

$$\int \frac{dv}{ a^{2} - v^{2}} = \ln \left ( \frac{ v + a}{ v - a } \right ) + c$$

extra square root, my mistake (senior moment). You can get this from parital fractions also....... an online list of integrals is at

http://www.mathwords.com/i/integral_table.htm

11. Jan 13, 2005

### dextercioby

Why didn't you post the correct formula??Where's that $\frac{1}{2a}$ ??

Daniel.

Last edited: Jan 13, 2005
12. Jan 13, 2005

### Dr Transport

I gave the general formula, just factor out the necessary terms and go from there.

13. Jan 13, 2005

### dextercioby

If this is the "general formula":
$$\int \frac{dv}{a^{2}-v^{2}}=\ln(\frac{v+a}{v-a})+C$$

,then it's deadly wrong since,for the formula to stand,it assumes a particular value for the integration constant (viz.
$$C=\ln(\frac{1}{2a})$$

),which would flagrantly contradict the rule of expressing the "family" of antiderivatives using ARBITRARY constants.

Daniel.

14. Jan 13, 2005

### Dr Transport

$$\int \frac{dv}{a^{2}-v^{2}}=\frac{1}{2a}\ln(\frac{v+a}{v-a})+C$$

constant factor included.......

15. Jan 13, 2005

### dextercioby

That's more like it... Now it looks goooooood... :tongue2:

Daniel.

16. Jan 13, 2005

### BigStelly

Hey guys this problem is getting to me and i want to be able to work at this to get v explicitly so im gonna keep at it.

$$\int \frac{dv}{mg-kv^{2}}=\int \frac{dt}{m}$$

following the general integration rule i took $$\sqrt{\frac{mg}{k}}$$ as a so to fit the form. Since k is a constant and is also a constant i rewrote that as k. I am not sure if all this simplifying of k is valid so if it isnt let me know.

The resulting integral is simply

$$\frac{1}{2k}\ln{\frac{v+k}{v-k}=\frac{t}{m}+C$$

Now let me know if this is all valid cuz i always make careless mistakes.

from there simplifying that one i get

$$\frac{v+k}{v-k}=Ce^{\frac{2kt}{m}}$$

Let me know if this is right?

17. Jan 13, 2005

### dextercioby

Pay attention with relabeling constants.In this case,i'm afraid you did it badly.You forkot one "k" (the "original one") when u factored it out from the denominator so u could put the latter as a difference of squares.

Daniel.

18. Jan 13, 2005

### BigStelly

Thanks man that helps alot.... but now im stuck on how to relabel so you can make the equation a bit nicer to work with any ideas? By the way i like your icon......

19. Jan 13, 2005

### dextercioby

Yes.Forget about the labeling this time.It's not helpful since it would mean the same number of constantsne the "k",one the relabeling (the sq.root),one from the integration;it would be better to carry arond that radical.

Thenx.I am an icon... :tongue2:

Daniel.

20. Jan 14, 2005

### BigStelly

Man leave it to a physicist to have an ego.... oh well if you know your stuff no point in pretending you dont right....?
Alright well so it should actually be....

$$\frac{1}{2\sqrt\frac{mg}{k}}\ln{\frac{v+sqrt\frac{mg}{k}}{v-sqrt\frac{mg}{k}}=\frac{t}{m}+C$$
Is this right? Any help appreciated......

Remember humility is good for even the most brilliant of minds....