Distance to reach terminal velocity

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Homework Help Overview

The discussion revolves around a physics problem concerning a 100 kg body falling and the concepts of terminal velocity, distance, and time to reach that velocity. Participants explore the mathematical relationships and equations involved in this context, particularly focusing on drag force and differential equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the nature of terminal velocity as an asymptote, questioning how to define "reaching" it. There are mentions of using differential equations and approximations to find time and distance. Some participants express uncertainty about their calculations and seek clarification on the use of variables versus numerical values.

Discussion Status

The conversation is active, with participants sharing their attempts and seeking feedback on their approaches. Some guidance has been offered regarding the use of integration and the importance of maintaining symbolic representations in equations. There is an ongoing exploration of assumptions related to drag coefficients and constants used in calculations.

Contextual Notes

Participants note the lack of specific values for the drag coefficient and question the appropriateness of certain constants used in their equations. There is a mention of the shape of the body affecting the drag coefficient, which adds complexity to the problem.

abdo799
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Homework Statement



A body with a 100 Kg of mass is falling , what's the distance and time to reach terminal velocity?

Homework Equations


a=dv/dt v=ds/dt drag force= 100v^2
F=ma

The Attempt at a Solution


I don't really need the numbers , just the expressions
I don't know how to write the equations so I am going to capture my attempt and post as photos. I don't know if I am right or wrong yet. The photos were mixed during the upload , the last photo is the first one and vice versa
 

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Sorry, x = infinity, t = infinity.

The terminal velocity is a finite number but the distance and time to reach it are asymptotes which are never completely reached.
 
rude man said:
Sorry, x = infinity, t = infinity.

The terminal velocity is a finite number but the distance and time to reach it are asymptotes which are never completely reached.

yea i know , but we can get an approximation , i could use excel to do it , but i saw this thread ( https://www.physicsforums.com/showthread.php?t=502933 ) he used differential equations to solve for time , so i thought maybe i could use it to solve for distance
 
Last edited:
abdo799 said:
yea i know , but we can get an approximation , i could use excel to do it , but i saw this thread ( https://www.physicsforums.com/showthread.php?t=502933 ) he used differential equations to solve for time , so i thought maybe i could use it to solve for distance

You tell me what constitutes "reaching terminal velocity", like for example 99% of terminal velocity, and maybe I can show you how to get that.
 
rude man said:
You tell me what constitutes "reaching terminal velocity", like for example 99% of terminal velocity, and maybe I can show you how to get that.
Thanks , i did a spreadsheet and i got the answers , my question is : is this calculus approach wrong?
 
abdo799 said:
Thanks , i did a spreadsheet and i got the answers , my question is : is this calculus approach wrong?
For the time, yes, though my preference would be to keep all the variables as symbols (m for mass, D for drag coefficient etc.) rather then inserting numerical values som early.
You can solve the time integral using partial fractions.
For the distance, the easiest way is to use dv/dt = (dv/ds)(ds/dt) = v dv/ds. That gives you a very easy integral.
 
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haruspex said:
For the time, yes, though my preference would be to keep all the variables as symbols (m for mass, D for drag coefficient etc.) rather then inserting numerical values som early.
You can solve the time integral using partial fractions.
For the distance, the easiest way is to use dv/dt = (dv/ds)(ds/dt) = v dv/ds. That gives you a very easy integral.

So integration does work ? thanks for that ultra-easy integral :thumbs: , the one i figured out was a nightmare
 
rude man said:
You tell me what constitutes "reaching terminal velocity", like for example 99% of terminal velocity, and maybe I can show you how to get that.

if i had a 50 m/s terminal velocity for example , if i put v= 49 m/s , i can work it out with calculus right?
 
abdo799 said:
if i had a 50 m/s terminal velocity for example , if i put v= 49 m/s , i can work it out with calculus right?

Right!
 
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  • #10
Thanks
 
  • #11
One last question before we call it a day , when i use differential equation calculator in wolfram , he used log instead of ln in the integration , he gave me t in terms of v . when he was trying to put v the subject ( to put v in terms of t ) he used e . How come?
 
  • #12
abdo799 said:
One last question before we call it a day , when i use differential equation calculator in wolfram , he used log instead of ln in the integration , he gave me t in terms of v . when he was trying to put v the subject ( to put v in terms of t ) he used e . How come?

What was your diff. equation?
 
  • #13
rude man said:
What was your diff. equation?

Int(t)=int(dv/(228-0.113v^2))
 
  • #14
abdo799 said:
Int(t)=int(dv/(228-0.113v^2))

OK, you mean dt = dv/(228 - 0.113 v2).

That "228" bothers me. I have g for that constant = 9.81 ms-2.
Also, you need to know the drag coefficient = drag force/v2. I saw nothing in your problem that gave that coefficient a number. Where did you get the number 0.113 from, which has to be the drag coefficient divided by the mass of the diver?
 
  • #15
rude man said:
OK, you mean dt = dv/(228 - 0.113 v2).

That "228" bothers me. I have g for that constant = 9.81 ms-2.
Also, you need to know the drag coefficient = drag force/v2. I saw nothing in your problem that gave that coefficient a number. Where did you get the number 0.113 from, which has to be the drag coefficient divided by the mass of the diver?

This is not really about skydiving, i am pretty sure the other calculation are correct, the body is teardrop shape which according to google has a coefficient of drag of 0.05
 
  • #16
OK, but I still think you should show how you got those two numbers: 228 and 0.113.

If you're sure they're right (and I don't think they are), then go ahead and show what your solution was for v(t) or t(v) or both.
 

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