Free fall velocity with air resistance/drag

  • #1
RJLiberator
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Homework Statement


I am analyzing free fall motion in my computer code class.
We haven't really discussed much about air resistance and it is a bit of a foreign topic for me.
I am searching the internet as much as I can for information on it, but would really appreciate talking to someone here regarding it.

What I've been giving is
F_d = -bv where b is a constant and v is velocity.
Teacher said to use our decay problem as an example code for this. Suggesting that I may use something of the form Ae^...

I'm just not sure how this all relates.


I search online and it seems like there are many many different types/examples of air resistance equations/examples.
 

Answers and Replies

  • #2
RJLiberator
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So I'm observing
v(t) = mg/b-(e^(-bt))*mg/b
But, in my problem there is no mass.
 
  • #3
RJLiberator
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Sorry to bump, this will be the last time.

I have to code this guy into a free fall problem, problem is, I haven't really worked with air resistance.

Many of the sites (youtube videos and the like) all have the drag force dealing with mass and other such things that I do not consider in my coding problem.

The teacher only gave us this problem:

dv/dt = a - bv

I tried solving this with my limited knowledge of differential equations and found
a = e^(-bt)+bv
But this doesn't make much sense to me as, how can a be a part of the velocity graph.
If acceleration mean the change in velocity maybe I could look at it as a time step? But then e^(-bt) is dependent on time anyway.
 
  • #5
RJLiberator
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Will do.

dv/dt = a - bv

Divide across by dv

1/dt = (a-bv)/dv
flip everything (inverse)
dt = dv/(a-bv)

integrate
t = -log(a-bv)/b

Multiply both sides by -b
-bt = log(a-bv)

exponentiate
e^(-bt) = a-bv

a = e^(-bt)+bv
 
  • #6
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Will do.

dv/dt = a - bv

Divide across by dv

1/dt = (a-bv)/dv
flip everything (inverse)
dt = dv/(a-bv)

integrate
t = -log(a-bv)/b

Multiply both sides by -b
-bt = log(a-bv)

exponentiate
e^(-bt) = a-bv

a = e^(-bt)+bv
You left out the constant of integration, required to satisfy the initial condition at t = 0.
 
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  • #7
RJLiberator
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At time = 0 there would be an initial velocity?

So would it just become a = e^(-bt)+bv-v_i
 
  • #8
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At time = 0 there would be an initial velocity?

So would it just become a = e^(-bt)+bv-v_i
The initial velocity is zero, and your equation doesn't satisfy that. The correct solution should be:

$$v=a\frac{(1-e^{-bt})}{b}$$
 
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  • #9
RJLiberator
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v = (a-ae^(-bt))/b

So this equation is not dependent on mass, which is good.
But, the problem I am having with this is it is dependent on acceleration and velocity.
I have the duty of coding a graph of the equation of velocity from a free falling object.
I have successfully made the graph for the case without air resistance.

So I figure the graph with air resistance shouldn't be too hard. It should be a simple extension.
When I look at this equation, I am left wondering how to incorporate it into my code.

v is what we are looking for.

1) b, I have no idea what b is. I know it's a constant that is dependent on the fluid it traverses, but how am I suppose to give it numerical value? Should I make it a user input and suggest some values for b?
b is determined by vterm = mg/b
But I don't have a mass, and what would be my terminal velocity? Don't have that either.

I suppose I can let the user input values for b, but I am struggling even finding realistic values to suggest.

2) And then there is acceleration. Can I represent that by a small change in time of the velocity? OR could that Just be gravity = 9.81?
 
  • #10
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I don't know how you want to handle b. It depends on how technically correct you want to be. Is this for a physics course or a math course?
You should have a = g in your equations. As far as terminal velocity is concerned, it is the value of v when t is infinite. What does your equation say?
 
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  • #11
RJLiberator
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I don't know how you want to handle b. It depends on how technically correct you want to be. Is this for a physics course or a math course?
You should have a = g in your equations. As far as terminal velocity is concerned, it is the value of v when t is infinite. What does your equation say?

I think the best route to go is to let b be user inputted at this point.

Thank you for the confirmation on a = g and understanding of terminal velocity.
When t is infinite, terminal velocity is a/b which is 9.81/b.

This is for a physics course entitled Mathematical and Computational methods for Physicists.

Your help here has been extremely important in my understanding of what I have to do.
 

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