1-d kinematic problem

[captain]

1. The problem statement, all variables and given/known data
A canonball is launched up in the air with velocity v_0. There is air resistance, which is equal to kmv (where k is some proportionality constant and m is the mass of the canonball) and is proportional to the canonball's velocity. How long does it take to reach its maximum height.


2. Relevant equations



3. The attempt at a solution

I got t=ln(g)/k as the time. I would just like to verify if that is the correct solution. g=earth's gravitational acceleration constant.

[Hootenanny]

Your answer is incorrect. Perhaps if you showed your working we could point you in the right direction...

[captain]

Your answer is incorrect. Perhaps if you showed your working we could point you in the right direction...

i realize that my units are off by a lot. i realized that i get the total acceleration to be
-kv+g=(dv)/(dt), but when i try and integrate it by (dv)/(-kv+g)=dt i get incorrect units. is the g supposed to stay on the other side of the equation. If so how can I integrate it? and after that I am not quite sure about how to use v_0 given to me?

[captain]

i would like to verify if this is correct. I checked to make sure if it is in the right units.

t=-ln[(v_0/g)k +1]/k

[learningphysics]

i would like to verify if this is correct. I checked to make sure if it is in the right units.

t=-ln[(v_0/g)k +1]/k

I'm getting exactly the same thing but without the minus sign.

I think the minus sign shouldn't be there because ln(positivenumber + 1) > 0... so with that minus sign there you'll get a negative time.

[captain]

i got it in terms of my integration bounds for the integral with dv. i set the bounds from v to 0

[learningphysics]

-kv+g=(dv)/(dt),

should you be using -kv - g... or are you using g = -9.8 instead of g = 9.8?

[captain]

i see i was doing it wrong the whole time

[captain]

i accidentally made the force in my free body diagram like this:

m(-a)=-f_drag-mg

i thought that since the acceleration was negative with respect to my reference frame, the -a was necessary when the right hand side of the equation took care of that.
thanks for your help.