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