Terminal Velocity proportional to the Drag Force 𝑚𝛾𝑣² in free fall

In summary: The terminal velocity equation as a function of the coefficient γ² is easy, but the problem is that once you substitute in the ##u## variable the equation becomes messy.
  • #1
Can anyone help me? I know that's wrong, but i don't know where.

Thanks for your attention so far.
 

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  • #2
What question do you wish to find out?
 
  • #3
@Victor Correa putting content in attachments is not acceptable. You need to post your content directly in the forum, using the PF LaTeX feature for equations. There is a LaTeX Guide link at the bottom left of the edit window when you are composing a post.
 
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  • #4
Obviously in #1 the OP looks for the solution of the equation of motion for free fall including air resistance,
$$\ddot{x}=\dot{v}=g-\gamma v^2.$$
This equation for ##v## can obviously solved by separation of variables,
$$\mathrm{d} t = \frac{\mathrm{d} v}{g-\gamma v^2}.$$
We need the integral
$$\int \mathrm{d} v \frac{1}{g-\gamma v^2} = \frac{1}{g} \int \mathrm{d} v \frac{1}{1-(\sqrt{\gamma/g}v)^2} = \frac{1}{\sqrt{g \gamma}} \text{artanh} \left (\sqrt{\frac{\gamma}{g}} v \right).$$
With the initial condition ##v(0)=0## we thus get
$$t=\frac{1}{\sqrt{g \gamma}} \text{artanh} \left (\sqrt{\frac{\gamma}{g}} v \right).$$
Solved for ##v## you get
$$v(t)=\sqrt{\frac{g}{\gamma}} \tanh (\sqrt{g \gamma} t).$$
The terminal velocity is
$$v_{\infty} = \sqrt{\frac{g}{\gamma}}.$$
 
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  • #5
Yeah,after the ##u## substitution things get messy in your solution.

Leave the LHS as:

$$ \frac{1}{\gamma} \int \frac{dv}{ \left(\sqrt{ \frac{g}{\gamma }}\right)^2 -v^2 }= \frac{1}{\gamma} \int \frac{A dv}{\sqrt{\frac{g}{\gamma}} + v}+\frac{1}{\gamma} \int \frac{B dv}{\sqrt{\frac{g}{\gamma}} -v}$$

and continue with the partial fraction decomposition from there.
 
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  • #6
The error you make is saying that the integral of du/(1-u) = ln(1-u)
 
  • #7
hutchphd said:
What question do you wish to find out?
The terminal velocity equation as a function of the coefficient γ²
 
  • #8
PeterDonis said:
@Victor Correa putting content in attachments is not acceptable. You need to post your content directly in the forum, using the PF LaTeX feature for equations. There is a LaTeX Guide link at the bottom left of the edit window when you are composing a post.
Oh, I didn't know. Sorry
 
  • #9
vanhees71 said:
Obviously in #1 the OP looks for the solution of the equation of motion for free fall including air resistance,
$$\ddot{x}=\dot{v}=g-\gamma v^2.$$
This equation for ##v## can obviously solved by separation of variables,
$$\mathrm{d} t = \frac{\mathrm{d} v}{g-\gamma v^2}.$$
We need the integral
$$\int \mathrm{d} v \frac{1}{g-\gamma v^2} = \frac{1}{g} \int \mathrm{d} v \frac{1}{1-(\sqrt{\gamma/g}v)^2} = \frac{1}{\sqrt{g \gamma}} \text{artanh} \left (\sqrt{\frac{\gamma}{g}} v \right).$$
With the initial condition ##v(0)=0## we thus get
$$t=\frac{1}{\sqrt{g \gamma}} \text{artanh} \left (\sqrt{\frac{\gamma}{g}} v \right).$$
Solved for ##v## you get
$$v(t)=\sqrt{\frac{g}{\gamma}} \tanh (\sqrt{g \gamma} t).$$
The terminal velocity is
$$v_{\infty} = \sqrt{\frac{g}{\gamma}}.$$
You are my hero !
 
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  • #10
Victor Correa said:
The terminal velocity equation as a function of the coefficient γ²
So that is easy indeed $$\ddot{x}=\dot{v}=g-\gamma v^2.$$ Just demand$$\dot{v}=g-\gamma v^2=0.$$
 
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  • #11
hutchphd said:
So that is easy indeed $$\ddot{x}=\dot{v}=g-\gamma v^2.$$ Just demand$$\dot{v}=g-\gamma v^2=0.$$
My problem is this 🤣
 
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  • #12
Victor Correa said:
My problem is this 🤣
Yeah, you had a bit of a "can't see the forest through the trees" issue.
 
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