Does time down equal time up with linear air resistance?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
10 replies · 4K views
ptk3
Messages
7
Reaction score
0
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)
 
Physics news on Phys.org
Well, what have you done so far?
 
I've solved the problem and would like to see if someone gets the same results.
 
Well, post your solution procedure, and we'll check if it is correct.
 
Let me finish writing it up and then I'll submit it in pdf form.
 
Here are the results for linear air resistance. The quadratic resistance results are being written up now. I hope I did the attachment correctly.
 
I don't think the file uploaded, so here's a copy:

http://caccmath.cacc.cc.al.us/air-resist.pdf
 
Last edited by a moderator:
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:
[tex]\frac{dv}{dt}=-g-\frac{k}{m}v (1)[/tex]
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: [tex]v=-\frac{mg}{k}[/tex]

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

Do you follow this?
 
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.
 
Welcome to PF!
 
So how did you get the TeX looking formulas in your post?