Comparing Times of Flight in a Linear Air Resistance Model

Click For Summary

Homework Help Overview

The discussion revolves around the times of flight for an object under a linear model of air resistance, specifically comparing the ascent and descent of the object. The original poster presents two differential equations representing the upward and downward motion, questioning whether the time taken to ascend equals the time taken to descend.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to define the times of flight for both upward and downward motion and questions the equality of these times. Some participants inquire about the original poster's progress and reasoning, while others suggest sharing the solution procedure for verification.

Discussion Status

The discussion includes attempts to clarify the reasoning behind the time comparison, with some participants expressing skepticism about the original poster's conclusion. There is an indication of guidance offered regarding the nature of air resistance and its effects on motion, but no consensus has been reached on the equality of the times.

Contextual Notes

Participants are navigating the complexities of air resistance in their analysis, with some noting potential flaws in reasoning. The original poster clarifies that this inquiry is not part of an assignment, which may influence the depth of exploration.

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.
 
  • #10
Welcome to PF!
 
  • #11
So how did you get the TeX looking formulas in your post?
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
Replies
6
Views
3K
Replies
2
Views
1K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
3K