# Motion of a body falling in resistive medium [closed]

Obtain the equation of motion for a particle falling freely under the action of gravity under a resistance which is directly proportional to the square root of velocity. Describe the motion and find the maximum velocity.

It appears to be a simple problem. We write $$m\frac{dv}{dt}=mg-b\sqrt v$$ $$\Longrightarrow \int_0^v\frac{dv}{mg-b\sqrt v}=\int_0^t\frac{dt}{m}$$ I am assuming that initial velocity is zero. Now, the right integral is straightforward. Let the left one be $$I$$ Apply substitution $$mg-b\sqrt v=u$$ Then $$\Longrightarrow\Big(\frac{mg-u}{b}\Big)^2=v$$ $$\Longrightarrow 2\Big(\frac{mg-u}{b}\Big)\Big(\frac{-1}{b}\Big)du=dv$$ Using it in the expression for $$I$$, we get, $$I=\int_{mg}^{mg-b\sqrt v}\frac{2(u-mg)}{b^2(u)}du$$ $$=\frac{2}{b^2}(-b\sqrt v)-\frac{2mg}{b^2}\ln\Big(\frac{mg-b\sqrt v}{mg}\Big)=\frac{t}{m}$$ $$\Longrightarrow\frac{2mg}{b^2}\ln\Big(\frac{mg}{mg-b\sqrt v}\Big)=\frac{t}{m}+\frac{2\sqrt v}{b}$$ $$\Longrightarrow \frac{mg}{mg-b\sqrt v}=e^{\frac{tb^2}{2m^2g}+\frac{b\sqrt v}{mg}}$$ $$\Longrightarrow \frac{mg-b\sqrt v}{mg}=e^{-\Big(\frac{tb^2}{2m^2g}+\frac{b\sqrt v}{mg}\Big)}$$ $$\Longrightarrow \sqrt v=\frac{mg}{b}\Bigg(1-e^{-\Big(\frac{tb^2}{2m^2g}+\frac{b\sqrt v}{mg}\Big)}\Bigg)$$ Now, this is an implicit function of $$\sqrt v$$, and I can't proceed to find the maximum value of $$\sqrt v$$ from here. What I think is, this appears to be an increasing function of time. Can I assume that the maximum value will occur at $$t\longrightarrow\infty$$? In that case, I think the $$\sqrt v$$ in the power will simply be ignored and maximum value will be $$\frac{mg}{b}$$.

Also, I thought of differentiating this expression with respect to time, and I got that $$\frac{d\sqrt v}{dt}$$ is never $$0$$.

## closed as off-topic by John Rennie, stafusa, Kyle Kanos, Jon Custer, ZeroTheHeroSep 6 at 21:11

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• Note that the question is using somewhat incorrect terminology. If the object is freely falling, then technically there cannot be a drag force. – Aaron Stevens Sep 5 at 13:37
• Unless my eyes are playing tricks, you missed a $u$ in the denominator of your integral when you make the substitution. You only have the $du$ term, but left out the 1/$u$, n'est ce pas? – Bill N Sep 5 at 17:11
• @BillN, yes I missed the term, corrected now. But that was while copying the partial solution from copy to here, so all the subsequent steps are correct. – Martund Sep 5 at 17:33
• From your first equation, the maximum velocity occurs when dv/dt = 0. v(max) =(mg/b)^2 – R.W. Bird Sep 5 at 18:58
• May I know the reason for downvote, dear downvoter? – Martund Sep 6 at 13:30

I will note that you essentially already have a solution to your problem, but I will supply you with some additional information and techniques that should help in problems like these.

Note that the question does not actually ask you to solve for $$v(t)$$. We can learn all that we need to learn from the differential equation. First, a change of variables makes this equation easier to deal with by introducing dimensionless variables $$x=\left(\frac{b}{mg}\right)^2\cdot v$$ $$\tau=\frac{b^2}{m^2g}\cdot t$$

Which changes the differential equation to be (check for yourself) $$\frac{\text dx}{\text d\tau}=1-\sqrt x$$

If we want to get an idea of the evolution of $$x$$, we can just plot $$\text dx/\text d\tau$$ as a function of $$x$$ Now notice that when $$x<1$$ we have $$\dot x>1$$. This means that $$x$$ will be "pushed" to larger values when $$x<1$$. Similarly, when $$x>1$$ we have $$\dot x<1$$. This means that $$x$$ will be "pushed" to smaller values when $$x>1$$.

Putting this all together, we see that $$x=1$$ is a stable equilibrium for this system. The system always tends to $$x=1$$. Furthermore, $$x$$ cannot move past $$x=1$$. So if we start at rest ($$x=0$$), then the maximum value $$x$$ will obtain is $$1$$ (although this doesn't tell us how long it will take to get there).

Furthermore, you can see how the system will approach $$x=1$$ based on the shape of the graph. For example, if $$\dot x$$ is larger, then $$x$$ is changing faster. If $$\dot x$$ is closer to $$0$$, then $$x$$ is not changing as quickly.

I will leave the specifics of how qualitatively $$x$$ changes with time, and how that relates back to $$v(t)$$ using the change of variables. What I do want you to realize is that you can learn a lot about a system without actually explicitly solving the differential equation. This is useful when the solution is messy (as you have seen).

• I don't have solution to my problem. – Martund Sep 5 at 14:37
• @Martund I meant you obtained a relation that solves the differential equation. Although it is very messy. – Aaron Stevens Sep 5 at 14:39
• Nice solution, but now when I have solved the differential equation, differentiating it back would give me the original differential equation. But when we differentiate and put $\frac{d\sqrt v}{dt}=0$, we don't get any solution, while for differential equation we directly get the maximum value of $v$. So, have I done something wrong or the maximum value will never be achieved? – Martund Sep 5 at 18:17
• @Martund I would assume the maximum value is not reached in finite time. This even happens for the usual drag forces like $-bv$. – Aaron Stevens Sep 5 at 18:24