Comparing Times of Flight in a Linear Air Resistance Model

In summary, the conversation revolves around a question regarding flight times under a linear model of air resistance. The problem is described, and the variables T_{up}, T_{down}, and h are defined. The question is whether T_{down} is equal to T_{up}, and the poster has solved the problem and wants to compare results. A solution method is requested and the poster provides a link to their work, which is then reviewed and corrected by another poster. The conversation ends with the original poster thanking the reviewer for their help and clarifying that it was not an assignment.
  • #1
ptk3
7
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)
 
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  • #2
Well, what have you done so far?
 
  • #3
I've solved the problem and would like to see if someone gets the same results.
 
  • #4
Well, post your solution procedure, and we'll check if it is correct.
 
  • #5
Let me finish writing it up and then I'll submit it in pdf form.
 
  • #6
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
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:
  • #8
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?
 
  • #9
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
Welcome to PF!
 
  • #11
So how did you get the TeX looking formulas in your post?
 

What is linear air resistance?

Linear air resistance is the force that opposes the motion of an object through air, causing it to slow down. It is directly proportional to the speed of the object.

How does linear air resistance differ from other types of air resistance?

Linear air resistance is a type of air resistance that is directly proportional to the velocity of the object. Other types of air resistance, such as quadratic air resistance, are proportional to the square of the velocity.

What factors affect the amount of linear air resistance experienced by an object?

The amount of linear air resistance experienced by an object depends on factors such as the speed of the object, the density of the air, the surface area of the object, and the shape of the object.

How does linear air resistance affect the motion of an object?

Linear air resistance acts in the opposite direction of the object's motion, causing it to slow down. As the object's speed decreases, the amount of air resistance also decreases until it reaches a state of equilibrium.

Can linear air resistance be reduced or eliminated?

Linear air resistance can be reduced by decreasing the speed of the object, changing the shape of the object to be more aerodynamic, or by decreasing the density of the air through which the object is moving. However, it cannot be completely eliminated as it is a natural phenomenon in our atmosphere.

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