1-D air resistance problem

[pimpalicous]

1. The problem statement, all variables and given/known data
A particle is projected vertically upward with a speed v0 under the influence of a drag force that is proportional to the particle's speed. As the particle falls back down, its terminal speed is vt.
(a) Find the time it takes to reach ymax

(b) Find ymax

(c)Show that your answers for a and b make sense by taking the limit where the drag force approaches zero. Do they reduce to the familiar eqns of 1st semester physics?

(d) evaluate a and b numerically for blah blah blah


2. Relevant equations
Fd=-k*m*v


3. The attempt at a solution

I got solutions for a and b pretty easily.

For a I wrote:
-mg-kmv=m\frac{dv}{dt}
I seperated and integrated

-\int dt=\int dv/(g+kv)
then
-t=\frac{1}{k}*ln[\frac{g+kv}{g+kv_{0}}]

I got a speed of
v=-\frac{-g}{k}+(\frac{g}{k}+v0)e-kt

I set this equal to zero and solved for t to get

\Large t=\frac{1}{k}*ln[\frac{g+v_{0}k}{g}]

For part b, I separated the v equation and integrated to find y

y(t)=-\frac{gt}{k}+(\frac{g}{k^{2}} +\frac{v_{0}}{k})[-e^(-kt) +1]

I plugged in the equation for t and got

ymax=-\frac{g}{k^2}*ln[1+\frac{v_{0}k}{g}]+(\frac{g}{k^2} + \frac{v_{0}}{k})(\frac{-g}{g+v_{0}k} +1)

I checked them in c by taking the limit of t and ymax as k approached zero.

The t checked perfectly. I got t=\frac{v_{0}}{g}.

The ymax didn't go so well.

ymax=-\frac{g}{k^2}*ln[1+\frac{v_{0}k}{g}]+(\frac{g}{k^2} + \frac{v_{0}}{k})(-g/(g+0) +1)

I used the approximation ln(1+x)=x-x^2/2 where v0k/g=x

ymax=-\frac{g}{k^2}[\frac{v_{0}k}{g}-(\frac{v_{0}^2*k^2}{2g^2}]

it reduced to

ymax=-\frac{v_{0}}{k}+\frac{v_{0}^2}{2g}

The second term is what I'm looking for but the first term obviously goes to infinity and can't be right. Where did I make a mistake?