# Homework Help: Linear air resistance

1. May 20, 2005

### ptk3

My question deals with times of flight under a linear model of air resistance. Let me describe the problem.

Up problem: v' = -g - kv, x(0) = 0, v(0) = v_0
Down problem: v' = -g + kv, x(0) = h, v(0) = 0

Define T_{up} and h (max. height) by the Up problem: x(T_{up}) = h, v(T_{up}) = 0

Define T_{down} by Down problem: x(T_{down}) = 0

Question: Does T_{down} = T_{up}? (prove/disprove)

2. May 20, 2005

### arildno

Well, what have you done so far?

3. May 20, 2005

### ptk3

I've solved the problem and would like to see if someone gets the same results.

4. May 20, 2005

### arildno

Well, post your solution procedure, and we'll check if it is correct.

5. May 20, 2005

### ptk3

Let me finish writing it up and then I'll submit it in pdf form.

6. May 20, 2005

### ptk3

Here are the results for linear air resistance. The quadratic resistance results are being written up now. I hope I did the attachment correctly.

7. May 20, 2005

### ptk3

I don't think the file uploaded, so here's a copy:

http://caccmath.cacc.cc.al.us/air-resist.pdf [Broken]

Last edited by a moderator: May 2, 2017
8. May 20, 2005

### arildno

That's some assignment..

There is, however, one MAJOR flaw in your reasoning, I'll give you the proper solution here:

1) Air resistance is ALWAYS in the opposite direction of the velocity!!!

Thus, in both the up&down case, with gravity acting in the negative direction, we have:
$$\frac{dv}{dt}=-g-\frac{k}{m}v (1)$$
Note that in the "down" case, v<0, that is, air resistance works upwards, as it should.

Note therefore, that the terminal velocity (i.e, no acceleration) will be: $$v=-\frac{mg}{k}$$

Rearranging (1), and multiplying with the integrating factor $$e^{\frac{kt}{m}}$$ we have:
$$\frac{d}{dt}(v(t)e^{\frac{kt}{m}})=-ge^{\frac{kt}{m}}$$
Or, integrating this from t=0 to an arbitrary time, we get:
$$v(t)e^{\frac{kt}{m}}-v(0)=-\frac{mg}{k}e^{\frac{kt}{m}}+g$$
That is:
$$v(t)=-\frac{mg}{k}(1-e^{-\frac{kt}{m}})+v(0)e^{-\frac{kt}{m}}$$

9. May 20, 2005

### ptk3

Thanks - the T_{up} = T_{down} was counter to my physical intuition, so I was suspicious. Thanks for your time. This was not an assignment.

10. May 20, 2005

### arildno

Welcome to PF!

11. May 20, 2005

### ptk3

So how did you get the TeX looking formulas in your post?