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I recently encountered a problem in which when throwing a ball upwards one has to determine whether it goes up or comes down faster when not only was gravity to be considered, but also air resistance with an acceleration of $mkv$ where $k$ is just some constant, $m$ is the mass, and $v$ is velocity. Since there is no horizontal component it is a one dimensional problem described by the differential equation: $$\frac{dv}{dt} = -g - mkv$$ Now as part of the problem the solution had to be found numerically for a particular example which is really quite trivial so I decided to attempt a proof for all values by finding a solution analytically. This is not for extra marks or really anything to do with the assessment but really just a bit of fun. So I solved the equation for velocity: $$\frac{dv}{dt} = -g - mkv$$ $$\frac{\frac{dv}{dt}}{-g-mkv} = 1$$ $$\int{\frac{1}{-g-mkv}dv}=\int{1 dt}$$ $$-\frac{1}{mk}\int{\frac{mk}{mkv + g}dv} = t + C_1$$ $$\int{\frac{mk}{mkv + g}dv} = -mkt - mkC_1$$ $$\ln{mkv + g} + C_2 = - mkt - mkC_1$$ $$\ln{mkv + g} = - mkt - mkC_1 - C_2$$ $$mkv + g = e^{- mkt - mkC_1 - C_2}$$ $$v = \frac{e^{- mkt - mkC_1 - C_2} - g}{mk}$$ $$v = e^{-mkt} \times \frac{e^{-mkC_1-C_1}}{mk} - \frac{g}{mk}$$ $$v=\lambda e^{-mkt} - \frac{g}{mk}$$ To find $\lambda$ in terms of starting velocity we can substitute in $v_0$ and $t = 0$: $$v_0 = \lambda e^{-mk\times0} - \frac{g}{mk}$$ $$\lambda = v_0 + \frac{g}{mk}$$ $$\therefore v = v_0e^{-mkt} + \frac{(e^{-mkt}-1)g}{mk}$$ Then I integrated to find displacement: $$r = \int{v dt}$$ $$r = \int{(v_0e^{-mkt} + \frac{(e^{-mkt}-1)g}{mk})dt}$$ $$r = -\frac{v_0}{mk}e^{-mkt} - \frac{e^{-mkt}g}{m^2k^2} - \frac{g}{mk}t + C$$ $$r = -\frac{v_0}{mk}e^{-mkt} - \frac{(e^{-mkt} + mkt)g}{m^2k^2} + C$$ To find $C$ in terms of initial displacement we can substitute in $r_0$ and $t = 0$: $$r_0 = -\frac{v_0}{mk}e^{-mk\times0} - \frac{(e^{-mk\times0} + mk\times0)g}{m^2k^2} + C$$ $$r_0 = -\frac{v_0}{mk} - \frac{g}{m^2k^2} + C$$ $$C = r_0 + \frac{v_0}{mk} + \frac{g}{m^2k^2}$$ So, $$r = -\frac{v_0}{mk}e^{-mkt} - \frac{(e^{-mkt} + mkt)g}{m^2k^2} + r_0 + \frac{v_0}{mk} + \frac{g}{m^2k^2}$$ $$r = (1-e^{-mkt})\frac{v_0}{mk} - \frac{(e^{-mkt} + mkt - 1)g}{m^2k^2} + r_0$$ $$r = (1 - (e^{-mkt} + mkt))\frac{g}{m^2k^2} + (1-e^{-mkt})\frac{v_0}{mk} + r_0$$ I then attempted to find the roots for both equations and found that whilst this is possible for velocity, displacement was impossible to solve analytically and when I put it into wolfram alpha it came back with the 'product log function' that from my admittedly limited investigation seems to be numerical. So, my question is:

Is an analytical proof that it takes longer for a ball to come down after it goes up in air resistance possible based on the conditions given by the question?

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    $\begingroup$ Suggestion to the question (v2): Replace mass $m$ with $1/m$, because a heavier ball is less affected by drag. $\endgroup$ – Qmechanic May 3 '15 at 7:14
  • $\begingroup$ That's actually a very valid point, I don't know why I didn't catch that. I'll let them know but they will probably only adjust it for next assessment, not this one, as this is a maths based course so it doesn't make a huge difference. Thanks anyway :) $\endgroup$ – J-S May 3 '15 at 21:44
  • $\begingroup$ Shouldn't the signs switch between up and down, gravity and drag do not always point into the same direction? Or will this be achieved through the constant? $\endgroup$ – rul30 May 4 '15 at 6:16
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Consider the total energy of the particle $$ E=\frac{mv^2}{2}+mgh $$ Then (assuming $k>0$): $$ \dot{E}=mv\dot{v}+mgv=mv[-g-mkv+g]=-m^2kv<0 $$ So when the particle is thrown up and returns to a given height it has less energy than when it was first there. Since the potential energies are the same the speed has fallen. That is it comes down slower than it went up, and there is no need to solve for the trajectory to prove this.

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  • $\begingroup$ I considered this approach but unfortunately the question is trajectory based. Whilst this is beyond the course and not part of marking, I'm looking for a proof based in the question's confines. If no trajectory based answer comes after a little while I will mark this as the answer as technically it does give analytical proof for all cases but it's not quite what I'm looking for. Thank you for putting the time into answering either way :) $\endgroup$ – J-S May 3 '15 at 21:41
  • $\begingroup$ The question as posted says nothing about being trajectory based. In fact it reads as though you are expected to prove the required result without finding the trajectory. $\endgroup$ – Conrad Turner May 4 '15 at 5:38
  • $\begingroup$ Yes I understand the question does not say anything about being trajectory based which is why I will acknowledge this as an answer if no others come. I miss phrased the question so yes this is an answer to the question I posted not the one I needed answered however the answer I'm looking for would be still be a correct answer to the posted question. So whilst you have answered the question I posted, you've not answered my question - through no fault of your own. I however do not want to mark it answered because I feel the answer I'm looking for may still be posted. $\endgroup$ – J-S May 4 '15 at 9:46
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You are making this rather hard for yourself.

You correctly solved for the velocity, which is of the form

$$v(t) = c_1 e^{-\alpha t} - \frac{g}{a}$$

where $a = mk$ and $c_1$ is found from the initial conditions. Integrating this expression should just give you

$$x(t) = -\frac{c_1}{a} e^{-at} - \frac{gt}{a}$$

I think that because you ended up splitting the coefficient for $c_1$ over the two terms, you also distributed $e^{-at}$ over two terms and it all became a mess. Exactly where you went wrong in your math I cannot tell (and we are explicitly not a "check my work" site) but I hope the above puts you back on track.

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  • $\begingroup$ thanks very much. Just thought I'd make it clear I'm definitely not looking for check my work it's just that the rules seemed very set on showing your work so I thought I'd show it all. Also I don't see how this can be solved for zero any easier? $\endgroup$ – J-S May 3 '15 at 12:10

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