# Quick Question regarding Air Resistance

#### Slimmy

I feel like this does not necessarily fit the format of a homework problem perse, but I am not sure where else to ask, but—

I was doing a thought experiment of a trajectory with air resistance (Air-resistance being proportional to the velocity with some constant), and was thinking to solve it doing force body diagrams. I was curious, would you have to do two force body diagrams in the y-direction due to air resistance changing direction? One for when the projectile is moving upward, and one when it is falling back down? Therefor you must ultimately integrate two different functions to find the time it takes? Adding the time it takes to get to y_max, then from there to y=0 to find the total time?

Or perhaps I am just unfamiliar with how integrals work entirely, and that when I integrate I am already taking into account that air resistance is a function of time, regardless of it changing directions in finality. Actually I think I just answered my question. :P

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#### Xyius

If you are modeling a projectile and want to take into account air resistance, you simply use Newtons Law.

$F=ma=mg-kv$
In other words
$mx''+kx'-mg=0$

This is a second order homogeneous differential equation which can easily be solved by assuming the form of the solution has the form $e^{rt}$ where r is a constant.

EDIT: If you are not familiar with solving differential equations then you wont be able to solve this any other way. (I do not know your mathematical background.)

#### Slimmy

If you are modeling a projectile and want to take into account air resistance, you simply use Newtons Law.

$F=ma=mg-kv$
In other words
$mx''+kx'-mg=0$

This is a second order homogeneous differential equation which can easily be solved by assuming the form of the solution has the form $e^{rt}$ where r is a constant.

EDIT: If you are not familiar with solving differential equations then you wont be able to solve this any other way. (I do not know your mathematical background.)
Yeah, I understand what you mean, I think I over-analyzed the problem trying to over-complicate things, but integration superseded.

So if I find a function of y, through this integration and such, that will be the function of y for the entire trajectory, correct? I am getting paranoid.

Thanks in return.

EDIT: Because I keep doing this thought experiment, where I just drop an object with air resistance. When I create a force body diagram, air resistance is in the opposite direction of gravity, so I want to make it:

$F=ma=-mg+kv$

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#### Xyius

Well firstly, don't feel bad about over complicating things. It shows that you are really thinking about the problem. Second, If you do not know differential equations or Calculus, then there isn't much point of me showing you how to solve this. (Again I do not know your mathematical background.)

But looking at this again, you can re-write it as a first order DE.

$mv'+kv-mg=0$ From here the solution is..
v=$\frac{gm}{k}\left[ 1+e^{-\frac{k}{m}t} \right]$

EDIT: And yes, those would be the correct direction of the forces.

#### Slimmy

Well firstly, don't feel bad about over complicating things. It shows that you are really thinking about the problem. Second, If you do not know differential equations or Calculus, then there isn't much point of me showing you how to solve this. (Again I do not know your mathematical background.)

But looking at this again, you can re-write it as a first order DE.

$mv'+kv-mg=0$ From here the solution is..
v=$\frac{gm}{k}\left[ 1+e^{-\frac{k}{m}t} \right]$

EDIT: And yes, those would be the correct direction of the forces.
Well if that equation I wrote is correct, then the sum of the forces changes, does it not? So while the projectile is going up it is:

$F=ma=-mg-kv$

Then when it comes back down (After Y_max) it would be:

$F=ma=-mg+kv$

So what I am asking is, would I have to integrate those two functions separately and find the time for both of them to find the total time? Or does this first differential equation/integration take into account of air resistance changing direction? Or perhaps because I am integrating air resistance as a function of time, then because it is already changing due to time, it takes into account when it goes positive as it falls back down...

I think I understand how to integrate this, but as far as the conceptual understanding of integration goes, I am a little bit shaky.

#### Xyius

The solution I wrote takes into account both directions. Again I ask, what is your mathematical background? You cannot just simply integrate the differential equation I posted above to get the solution. You must use the method for solving a homogeneous differential equation.

If you want to read up about how to solve these types of equations, here is a link.
http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

EDIT: I gave the wrong link at first, it should be fine now.

#### Slimmy

The solution I wrote takes into account both directions. Again I ask, what is your mathematical background? You cannot just simply integrate the differential equation I posted above to get the solution. You must use the method for solving a homogeneous differential equation.

If you want to read up about how to solve these types of equations, here is a link.
http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

EDIT: I gave the wrong link at first, it should be fine now.
Thank you very much for your help!

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