I Terminal Velocity proportional to the Drag Force š‘šš›¾š‘£Ā² in free fall

AI Thread Summary
The discussion centers on solving the equation of motion for free fall with air resistance, represented by the equation ddot{x}=dot{v}=gāˆ’Ī³v². Participants clarify that the solution involves separating variables and integrating, leading to the expression for velocity as a function of time, v(t)=sqrt{g/γ}tanh(sqrt{gγ}t). The terminal velocity is derived as vāˆž=sqrt{g/γ}. There is an emphasis on using the PF LaTeX feature for posting equations directly in the forum, and a reminder about the importance of proper integration techniques. The conversation highlights the challenges and common errors in understanding the relationship between drag force and terminal velocity in free fall scenarios.
Victor Correa
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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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What question do you wish to find out?
 
@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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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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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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The error you make is saying that the integral of du/(1-u) = ln(1-u)
 
hutchphd said:
What question do you wish to find out?
The terminal velocity equation as a function of the coefficient γ²
 
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
 
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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