1-d kinematic problem

1. Oct 4, 2007

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.

2. Oct 4, 2007

Hootenanny

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

3. Oct 4, 2007

captain

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?

4. Oct 4, 2007

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

5. Oct 4, 2007

learningphysics

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.

6. Oct 4, 2007

captain

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

7. Oct 4, 2007

learningphysics

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

8. Oct 4, 2007

captain

i see i was doing it wrong the whole time

9. Oct 4, 2007

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.