Distance to reach terminal velocity

[abdo799]

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

 

[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.

 

[abdo799]

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

 

[rude man]

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.

 

[abdo799]

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?

 

[haruspex]

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.

 

[abdo799]

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

 

[abdo799]

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?

 

[rude man]

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!

 

[abdo799]

Thanks

 

[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?

 

[rude man]

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?

 

[abdo799]

What was your diff. equation?

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

 

[rude man]

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?

 

[abdo799]

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

 

[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.