# Homework Help: ODE problem

1. Jan 27, 2008

### bndnchrs

[SOLVED] ODE problem

1. The problem statement, all variables and given/known data
A situation in which the air resistance is proportional to the velocity of an object squared.

Object dropped off of a building with height 100m.

F = -mg + Fr Fr=.5*cw*p*A*v^2

cw=.5 p = air density = 1?

2. Relevant equations

I need to come about with a solution for y(t)

3. The attempt at a solution

I've used Mathcad to help me with a problem when the velocity is proportional to only the velocity linearly, using equations derived analytically.

I just don't know where to begin with this one. I can't derive it analytically to obtain an expression for v(t) so that I can throw that into my runga-kutta solver along with the force expression as the two time derivatives.

I have tried using ODE solver, but it fails to produce anything. It works for two iterations, up to t=2, then quits on me when I try to obtain more results.

I believe my best approach is to solve this analytically, but I can't do it, my prof. did something strange with letting the variable be v^2, but the exercize he did this in was a few weeks ago, and foggy in my mind. Thanks for any help!

2. Jan 27, 2008

### Rainbow Child

From Newton's 2nd Law

$$m\,\frac{d^2\,y}{d\,t^2}=m\,g-k\,v^2\Rightarrow \frac{d^2\,y}{d\,t^2}=g-\frac{k}{m}\,v^2 \quad (1)$$

where $k$=constant. You can write the accerelation as

$$\frac{d^2\,y}{d\,t^2}=\frac{d\,v}{d\,t}=\frac{d\,y}{d\,t}\frac{d\,v}{d\,y}=v\,\frac{d\,v}{d\,y}$$

$$v\,\frac{d\,v}{d\,y}=g-\frac{k}{m}\,v^2\Rightarrow \frac{v\,d\,v}{g-\frac{k}{m}\,v^2}=d\,y$$

Integrate the above equation to obtain $v=v(y)$. Then integrate
$$v=\frac{d\,y}{d\,t}$$
to obtain $y(t)$.

3. Jan 27, 2008

### bndnchrs

ahh... thank you very much. Didn't think about the change of base.

4. Jan 27, 2008

### Rainbow Child

Integrating
$$\frac{v\,d\,v}{g-\frac{k}{m}\,v^2}=d\,y$$
yields

$$v=-\sqrt{-\frac{m\,g}{k}+c_1\,e^{-2k\,y/m}}\Rightarrow \int\frac{d\,y}{\sqrt{-\frac{m\,g}{k}+c_1\,e^{-2k\,y/m}}}=-\int d\,t$$

For the first integral make the change of variables

$$\sqrt{-\frac{m\,g}{k}+c_1\,e^{-2k\,y/m}}=z,\, d\,y=-\frac{z}{g+\frac{k}{m}\,z}\,d\,z$$

Last edited: Jan 27, 2008
5. Jan 27, 2008

### bndnchrs

you are a lifesaver. Thanks a ton.