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