# 1-d kinematic problem

## Homework Statement

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.

## 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.

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

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?

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
Homework Helper
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.

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

learningphysics
Homework Helper
-kv+g=(dv)/(dt),
should you be using -kv - g... or are you using g = -9.8 instead of g = 9.8?

i see i was doing it wrong the whole time

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.