- #1
The terminal velocity equation as a function of the coefficient γ²hutchphd said:What question do you wish to find out?
Oh, I didn't know. SorryPeterDonis 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.
You are my hero !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
$$v_{\infty} = \sqrt{\frac{g}{\gamma}}.$$
So that is easy indeed $$\ddot{x}=\dot{v}=g-\gamma v^2.$$ Just demand$$\dot{v}=g-\gamma v^2=0.$$Victor Correa said:The terminal velocity equation as a function of the coefficient γ²
My problem is thishutchphd said:So that is easy indeed $$\ddot{x}=\dot{v}=g-\gamma v^2.$$ Just demand$$\dot{v}=g-\gamma v^2=0.$$
Yeah, you had a bit of a "can't see the forest through the trees" issue.Victor Correa said:My problem is this![]()