# 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....