# Linear drag

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1. Nov 1, 2014

### Patrikp

1.Small ball with mass m falls from hight h. Find how velocity, acceleration and height depend on time. Air resistance is linear and depends on velocity $F = kv$.

2. Air resistance $F=kv$
where $k$ is constant.

3. I do not know where to go after $ma = mg - kv$.

2. Nov 1, 2014

### Staff: Mentor

a = dv/dt and v = ds/dt
Do you know anything about solving differential equations?

3. Nov 1, 2014

### Patrikp

That is the problem... I know that a= dv/dt but where do I go then? I tried to integrate but got nowhere.

4. Nov 1, 2014

### Staff: Mentor

Show me what you did...

5. Nov 1, 2014

### Patrikp

I wrote it like $(dv)/(dt) + v(k/m)= g$
And then multiplied by $e^{∫(k/m)dt}$
to get
$v= e^{-(∫(k/m)dt)}(∫ge^{∫(k/m)dt}dt + c)$
and that gave something wrong.

6. Nov 1, 2014

### Staff: Mentor

Why don't you simplify $e^{\int (k/m)dt}$? There's no point in leaving it like this.

7. Nov 1, 2014

### Patrikp

I simplified but did not write it here... Anyways the result i should get is $v(t)= ((mg)/k )(1-e^{-((kt)/m))})$ and I cannot get it. Do you see mistake ??

8. Nov 1, 2014

### Staff: Mentor

Your work should include a line like this:
$$ve^{(k/m)t} = g\int e^{(k/m)t}dt = g(m/k)e^{(k/m)t} + C$$
Now solve for v.

9. Nov 1, 2014

### Patrikp

Thanks.
But I still can't get right result? And I always get that C hanging there?

10. Nov 1, 2014

### Staff: Mentor

What do you get when you solve for v?

11. Nov 1, 2014

### Patrikp

$v= (g/m)k + Ce^{-((kt)/m)}$

12. Nov 1, 2014

### Staff: Mentor

How did g(m/k) become (g/m)k? The other term looks fine.

13. Nov 1, 2014

### Patrikp

It's typo. I dont know why the solution is wrong then.
Thanks for help!

14. Nov 1, 2014

### Staff: Mentor

There are initial conditions that aren't being used -- at t = 0, s(0) = h and v(0) = 0. The initial velocity isn't given explicitly, but I am inferring it from the statement in the 1st post - small ball falls from height h.

Use v(0) = 0 to get rid of the constant C. Otherwise, the solution is correct, which I've checked, and you should as well.