Distance to reach terminal velocity

1. Dec 30, 2013

abdo799

1. The problem statement, all variables and given/known data

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

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

3. 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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2. Dec 30, 2013

rude man

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.

3. Dec 30, 2013

abdo799

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: Dec 30, 2013
4. Dec 30, 2013

rude man

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.

5. Dec 30, 2013

abdo799

Thanks , i did a spreadsheet and i got the answers , my question is : is this calculus approach wrong?

6. Dec 31, 2013

haruspex

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.

7. Dec 31, 2013

abdo799

So integration does work ??? thanks for that ultra-easy integral :thumbs: , the one i figured out was a nightmare

8. Dec 31, 2013

abdo799

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?

9. Dec 31, 2013

rude man

Right!

10. Dec 31, 2013

abdo799

Thanks

11. Dec 31, 2013

abdo799

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. Dec 31, 2013

rude man

13. Dec 31, 2013

abdo799

Int(t)=int(dv/(228-0.113v^2))

14. Dec 31, 2013

rude man

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. Dec 31, 2013

abdo799

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. Dec 31, 2013

rude man

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.