Trying to use Desmos to curve-fit some positional data of a ball bearing falling through water, and I want that curve to be relevant to $F_D=-kv^2$. So by mixing that with Newton's 2nd law, I got this equation: $$ mg-k\left(\frac{dx}{dt}\right)^2=m\frac{d^2x}{dt^2} $$ Where $m$, $g$, and $k$ are positive, non-zero constants, and for all $t\geq0$, $x(t)\geq0$, $x'(t)>0$
But I can't see how to solve the thing for $x$.
Any ideas?
Hint: Start by solving a simpler equation:$$m\frac{dv}{dt}=mg-kv^2$$ Then integrate to get $x$. Let me know if you still have problem with this.
Edit: $$\frac{dv}{dt}=g\left(1-\frac k{mg}v^2\right)$$ Note that you can define the terminal velocity when the acceleration is zero, so $$v_T^2=\frac{mg}k$$ Then $$\int\frac{dv}{1-\frac{v^2}{v_T^2}}=g\int dt$$ Integrating, you get $$v_T\mathrm{arctanh}\frac{v}{v_T}=gt+C$$ From here, I would need the initial velocity to get $C$. Let's assume that $v(0)=0$. Then $C=0$ and $$v=v_T\tanh\frac{gt}{v_T}$$ Integrating $\tanh$ is easy (it's $\sinh$ divided by $\cosh$). Assuming $x(0)=0$, you get $$x(t)=\frac{v_T^2}g\ln\cosh\frac{gt}{v_T}$$