Solve k in this ODE: mv′(t)=mg−kv^2(t)

Question

I have this question about finding the air resistance using the following differential equation:

*mv′(t)=mg−k(v^2)(t)*

where m is the mass of the downward-falling object, gravitational acceleration is g= 9.82, and k is a constant which describes the air resistance. According to the question, after many trials it was deduced that the constant fall speed is : 3.7√(m/A)m/s where m is mass in kg and A is the cross-section area in m^2.

So what is the value of k?

The only thing I know is that this equationis seprable but I am completely lost in this problem so any hint will be appreciated. Where should I start?

Thank you so much!

Answer 1

As a general rule, don't try to solve a differential equation if you don't need to.

When the object is falling at its constant fall speed, its acceleration (by definition) is zero. You've been told $v$ and $g$. You now know $v'$. The only unknowns remaining in the question are $k$ and $m$. In principle, that's a problem if you're expecting a fixed number for $k$; the correct response to that issue is usually to just try to solve for $k$ in terms of $m$, and hope the $m$'s cancel out.

Answer 2

We first look at the stationeries states: $v’(t)=0\Rightarrow k{v_0}^2=mg$. You can from here deduce $k$ since the constant fall speed $v_0$ is given.

Resolution: Now we put this expression into ODE. It comes: $mdv=k({v_0}^2-v^2)dt\Rightarrow \frac{dv}{v^2-v_0^2}=-\frac{k}{m}dt$ then: $$\int_{0}^{v}{\frac{dv}{v^2-v_0^2}}dv=\int_{0}^{\tau}{\frac{-k}{m}}dt$$ $\Rightarrow$ $$\frac{1}{v_0}arctanh {({\frac{v}{v_0}})}=\frac{k}{m}t$$ $\Rightarrow$ $$v=v_0\tanh(\frac{kv_0}{m}t)$$