Particle moving under gravity

1. Nov 11, 2009

leviathanX777

1. A particle moves vertically under gravity and a retarding force proportional to the square of its velocity. If v is upward or downward speed, shot that a = +/-g -kv^2, where k is a constant. If the particle is moving upwards, show that its position at time t is given by;

z = z0 +(1/k)lncos[rootgk (t0-t)] where z0 and t0 are integration constants.

2. Relevant equations

3. I proved the a = +/-g -kv^2 but I spent almost an hour looking at the next part with no results. I integrated twice to get x(t) = -gt^2/2 - kv^2t^2/2 + z0 + t0

I have no idea what to do after this, I mean I dunno how you bring the cos into the problem. This hint is also given; lncosx = -1/2 ln(1 + tan^2x)

Cheers

2. Nov 11, 2009

gabbagabbahey

How exactly are you integrating $-kv^2$ with respect to time, if $v(t)$ is some unknown function of time?

Instead, realize that $a(t)=\frac{dv}{dt}=-g-kv^2$...which gives you a separable 1st order ODE....surely you've come across those before?

3. Nov 11, 2009

leviathanX777

Oh yes sorry I thought for some reason you could just integrate it with respect to time. But I don't see how changing it into a first order would help with the problem.

4. Nov 11, 2009

gabbagabbahey

Have you not studied 1st order separable ODEs in any of your calculus courses yet?

5. Nov 11, 2009

leviathanX777

Yes we did them last year but I don't see how doing that will bring tan into the question.

6. Nov 11, 2009

gabbagabbahey

Solve the ODE for v(t) and then use the fact that $v(t)=\frac{dz}{dt}$ to solve your resulting equation for z(t)...

7. Nov 11, 2009

leviathanX777

Alright cheers, will try that.

8. Nov 11, 2009

leviathanX777

Did it but I knew that a(t) = dv/dt anyway
Still got the same results when I integrated

9. Nov 11, 2009

gabbagabbahey

If you show me your steps, I can point out your error(s)...

10. Nov 11, 2009

leviathanX777

a(t) = dv/dt = -g-kv^2
integrate, v(t) = -gt -kv^2t + t0

integrate again to achieve z(t):

Z(t) = -gt^2 - kv^t^2 + z0

11. Nov 11, 2009

gabbagabbahey

No, you are making the same mistake you did in your first attempt...

$$\int kv^2dt=kv^2t$$

only if v(t) is a constant....which it clearly isn't.

Open up your old calc text and review how to solve separable 1st order ODEs.

12. Nov 12, 2009

leviathanX777

Ah yes, sorry I totally forgot what to do, didn't realize. Here's my solution;

dv/dt = -g -kv^2
integral(dv/-g-kv^2) = integral(dt)
integral[(-1/g+kv^2)(dv)] = integral(dt)
-1/k*integral[v^2 + (root(g/k)^2)] = integral(dt)
Used math's tables; integral(1/x^2 + a^2) = 1/a inversetan x/a

Cancelled terms etc, v = dz/dt

Got z;
z = zo - root(g)/root(k) ln [cos(root(gk))(to-t)]

z = zo + (1/k) ln [cos(root(gk))(to-t)]

Any ideas?

13. Nov 12, 2009

gabbagabbahey

Looks like you might have forgotten the factor of $1/a$ from

$$\int\frac{1}{v^2+a^2}dv=\frac{1}{a}\tan^{-1}\left(\frac{v}{a}\right)+C$$

....but unless you post the rest of your steps I can't tell for sure.

14. Nov 12, 2009

leviathanX777

I didn't;
when applying that inverse tan I got;

-1/k[root(k)/root(g) inversetan(v*root(k)/root(g))] = T-To

15. Nov 12, 2009

gabbagabbahey

Okay, and what did you get when you solved that equation for v(t)?

16. Nov 12, 2009

leviathanX777

v(t) = root(g)/root(k) * tan (root(gk)*T1)

where T1 = T-T0

17. Nov 12, 2009

gabbagabbahey

What happened to the negative sign you had before?

18. Nov 12, 2009

leviathanX777

I just changed it mutiplied across by -1 and got; -T + T0 on the right hand side which I called T1

19. Nov 12, 2009

gabbagabbahey

Okay, so now you have

$$v(t)=\frac{dz}{dt}=\sqrt{\frac{g}{k}}\tan\left(\sqrt{gk}(t_0-t)\right)$$

Now use the variable substitution $u=\sqrt{gk}(t_0-t)$ to integrate the RHS....