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- Thread starter Victor Correa
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hutchphd

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What question do you wish to find out?

- #3

PeterDonis

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$$\ddot{x}=\dot{v}=g-\gamma v^2.$$

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

$$v_{\infty} = \sqrt{\frac{g}{\gamma}}.$$

- #5

erobz

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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.

- #6

mjc123

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The error you make is saying that the integral of du/(1-u) = ln(1-u)

- #7

Victor Correa

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The terminal velocity equation as a function of the coefficient γ²What question do you wish to find out?

- #8

Victor Correa

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Oh, I didn't know. Sorry

- #9

Victor Correa

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You are my hero !

$$\ddot{x}=\dot{v}=g-\gamma v^2.$$

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

$$v_{\infty} = \sqrt{\frac{g}{\gamma}}.$$

- #10

hutchphd

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So that is easy indeed $$\ddot{x}=\dot{v}=g-\gamma v^2.$$ Just demand$$\dot{v}=g-\gamma v^2=0.$$The terminal velocity equation as a function of the coefficient γ²

- #11

Victor Correa

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My problem is this 🤣So that is easy indeed $$\ddot{x}=\dot{v}=g-\gamma v^2.$$ Just demand$$\dot{v}=g-\gamma v^2=0.$$

- #12

erobz

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Yeah, you had a bit of a "can't see the forest through the trees" issue.My problem is this 🤣

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