A body falls through a medium I'm hating these variable resistance questions.
A body of mass $m$ falls from rest in a medium that produces a resistance of magnitude $m\cdot k \cdot v$. where $k$ is a constant, where the speed of the particle is $v$. Show that when the body has reached a speed $V$ it will have fallen for a time
$$\frac{1}{k} \ln \left(\frac{g}{g-k\cdot V}\right) \; .$$
Help much appreciated.
 A: Well the force on the particle due to gravity is $mg,$ where $g$ is the acceleration due to gravity, so the net downward force on the particle is $m(g-kv).$ So its acceleration is
$g-kv.$ So you need to solve $dv/dt = g-kv$ given the initial conditions. That is
$$\int_0^V \frac{ \text{d}v}{g-kv} = \int_0^T \text{d}t.$$
A: Okay here is my noob way of solving this  (I only learned this less than a year ago but I have forgotten everything and semester starts in a few weeks and I'll probably be required to do this stuff EEEEEEEEEEEEEEEEEEEEEEP),
The forces on the "body" are the resistance and its weight (gravity) as far as I can tell from your description
So 
$$ \vec{F}_{net} = m\mathbf{a} =mg-mk\mathbf{v} $$
which can be simplified to
$$  \mathbf{a} = g-k\mathbf{v}  $$
which can be written in a slightly different form as because acceleration is derivative of velocity
$$  v'(t)+k v(t)=g  $$
This looks type of equation looks familiar to me and can be solved as it is a "linear first order differential equation"
Now here is where I may start to make errors so please forgive me
First we multiple everything an "integrating factor" equal to $e^{\int f(t)\,dt}$ where $f(t)=k,$ to make it all work so
$$e^{\int f(t)\,dt}=e^{\int k\,dt}=e^{kt}$$  
We don't need a "plus C" (I'm not entirely clear on this but I don't think it makes a difference to anything in the long run, SOMEONE PLEASE CONFIRM)
So when we multiply everything by $e^{kt} $
$$  v'(t)e^{kt}+k v(t)e^{kt}=ge^{kt} $$
which is "equivalent" to this (look at the product rule)
$$  (v(t)e^{kt})'=ge^{kt} $$
Now if we integrate both sides
$$  v(t)e^{kt}={\int ge^{kt}\,dt}$$
$$  v(t)e^{kt}=\frac{ge^{kt}}{k}+C$$
$$  v(t)=\frac{C}{e^{kt}}+\frac{g}{k}$$
To get rid of the C, you said it starts from rest so $v(0)=0$
$$  v(0)=0=\frac{C}{e^{k0}}+\frac{g}{k} = \frac{C}{1}+\frac{g}{k}$$
$$ 0= C+\frac{g}{k}$$
$$ C=-\frac{g}{k}$$
So 
$$  v(t)=\frac{-\frac{g}{k}}{e^{kt}}+\frac{g}{k}$$
and I would simplify this to
$$  v(t)=g\frac{1-e^{-kt}}{k}$$
Now Wolfram Alpha would have done this for you
Now to the second part of your question when v(t)=V what is t?
$$  v(t)=V=g\frac{1-e^{-kt}}{k}$$
$$  \frac{kV}{g}={1-e^{-kt}}$$
$$  e^{-kt}=1-\frac{kV}{g}$$
$$  e^{-kt}=\frac{g-kV}{g}$$
$$  e^{kt}=\frac{g}{g-kV}$$
$$  kt=\log_e {\frac{g}{g-kV}}$$
$$  t=\frac{1}{k}\log_e ({\frac{g}{g-kV}})$$
Wolfram Alpha would do this too but is down for me now
I hope this is correct
