So for my latest physics homework question, I had to derive an equation for the terminal velocity of a ball falling in some gravitational field assuming that the air resistance force was equal to some constant c multiplied by $v^2.$
So first I started with the differntial equation:
$\frac{dv}{dt}=-mg-cv^2$
Rearranging to get:
$\frac{dv}{dt}=-\left(g+\frac{cv^2}{m}\right)$
From here I tried solving it and ended up with:
$\frac{\sqrt{m}}{\sqrt{c}\sqrt{g}}\arctan \left(\frac{\sqrt{c}v}{\sqrt{g}\sqrt{m}}\right)+C=-t$
I rearranged this to get:
$v\left(t\right)=\left(\frac{\sqrt{g}\sqrt{m}\tan \left(\frac{\left(-C\sqrt{c}\sqrt{g}-\sqrt{c}\sqrt{g}t\right)}{\sqrt{m}}\right)}{\sqrt{c}}\right)$
In order to calculate the terminal velocity I took the limit as t approaches infinity:
$\lim _{t\to \infty }\left(\frac{\sqrt{g}\sqrt{m}\tan \:\left(\frac{\left(-C\sqrt{c}\sqrt{g}-\sqrt{c}\sqrt{g}t\right)}{\sqrt{m}}\right)}{\sqrt{c}}\right)$
This reduces to:
$\frac{\sqrt{g}\sqrt{m}\tan \left(\infty \right)}{\sqrt{c}}$
The problem with this is that tan $(\infty)$ is indefinite.
Where did I go wrong? Could someone please help properly solve this equation.
Cheers, Gabriel.
[Math] How to derive an equation for terminal velocity assuming air resistance is some constant multiplied by the square of velocity
infinityordinary differential equationsphysics
Best Answer
Write the differential equation as a rate of change of velocity with respect to just aerodynamic drag. Then solve for the time it takes for the drag to equal $mg$.
$$\frac{dV}{dt} = \frac{cv^2}{m}$$ $$\frac{v^{-2}}{c}dV = \frac{dt}{m}$$ $$-\frac{1}{cv} = \frac{t}{m} + C$$ Assuming $t=0, v=0$ then.......
$$v = -\frac{m}{ct}$$ When $cv^2 = -mg, v = -\sqrt{\frac{gm}{c}}$ $$-\sqrt{\frac{gm}{c}} = -\frac{m}{ct}$$ $$t = \frac{m}{c\sqrt{\frac{gm}{c}}}$$ Substituting back.......$$v = \sqrt{\frac{gm}{c}}$$ Does this seem reasonable? Assume $c = .5\cdot C_d\cdot \rho\cdot A = .5\cdot 0.3\cdot 1.225\cdot 0.1 = 0.018$ and $m = 0.5\ kg$
$$v = \sqrt{\frac{9.8\cdot 0.5}{0.018}} = 16.5\ m/s$$
Thinking about this it would have been easier just to set $cv^2 = mg$ to get $$v = \sqrt{\frac{gm}{c}}$$