Mathematics of simplified air resistance 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?
 A: 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.
A: 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.
