The full question is:
A particle of mass m is projected vertically upwards through a resistive medium with initial speed $u$. It experiences a resistive force of $mkv$. and a weight force $mg$. When the same particle lands on the ground, it has speed $w$. Prove the $u>w$.
I have been able to find expressions for time and height of the upwards a downwards journeys as functions of $u$ and $w$ by manipulating the differential equations for acceleration.
On the upwards journey, taking the upwards direction as positive:
$a=-g-kv$
$T=\frac{1}{k}ln\left(\frac{g+ku}{g}\right)$
$h=\frac{1}{k}u+\frac{g}{k^2}ln\left(\frac{g}{g+ku}\right)$
On the downwards journey, taking the downwards direction as positive:
$a=g-kv$
$T=\frac{1}{k}ln\left(\frac{g}{g-kw}\right)$
$h=\frac{-1}{k}w+\frac{g}{k^2}ln\left(\frac{g}{g-kw}\right)$
I have tried various things like arguing that the time down was greater than time up and thereby creating an inequality. I have also tried equation $h$. However, I have not been able to arrive at the desired result.
Best Answer
I appreciate this might not be the way you would expect this to be done, but basically I use the fact that all that the drag does is dissipate energy, so the sum of the potential and kinetic energy must go down.
Let's agree to measure everything (displacement, velocity, acceleration) as positive on the way "up": then $a=-g-kv$ all the time, no matter whether the object is going up or down. Namely, going "up", $v$ is positive and the drag acts "down" so you take both $g$ and $kv$ with the negative sign. On the way "down", the drag acts "up" but as $v$ is negative we have to again take both $g$ and $kv$ with the negative sign.
Now, we will differentiate the sum $E_k+E_p$ of the potential and kinetic energy by time: you have:
$$\begin{array}{rcl}\frac{d}{dt}(E_p+E_k)&=&\frac{d}{dt}\left(mgh+\frac{mv^2}{2}\right)\\&=&mgv+mva\\&=&mgv+mv(-g-kv)\\&=&-mkv^2\\&\le&0\end{array}$$
and, in fact, you would have $<0$ at any point where the particle is moving. In other words, the derivative by time of the sum of the potential and kinetic energy is negative, so the sum at the end (highest point) must be smaller than at the start. Mathematically:
$$(E_{p_\text{end}}+E_{k_\text{end}})-(E_{p_\text{start}}+E_{k_\text{start}})=\int_0^T \frac{d}{dt}(E_p+E_k)dt>0$$
as the function under the integral sign is nonnegative (and actually positive at any time the particle is moving).
So, at the end of the journey, $E_{p_\text{start}}=E_{p_\text{end}}$ so $E_{k_\text{start}}>E_{k_\text{end}}$ i.e. $\frac{mv_\text{start}^2}{2}>\frac{mv_\text{end}^2}{2}$ i.e. $|v_\text{start}|>|v_\text{end}|$.