# Homework Help: Free fall w/ Air Resistance

1. Feb 15, 2012

### jfrank1034

A ball of mass m is released from rest at x=0. Air resistance is expressed as R = kv2 , where k is a positive constant and v denotes velocity.

Derive an expression for the speed in terms of the distance x that it has fallen.
Identify the terminal v.

I know I have to start with ma = mg - kv2 and integrate but I can't seem to separate the variables right to get it as function of x or distance (even though the problem has a picture of a vertical fall)

2. Feb 15, 2012

### LawrenceC

Hint: Express the 'a' in your equation differently.

3. Feb 15, 2012

### jfrank1034

Okay so I have mv(dv/dx) = mg - kv2 , and divide both sides by the (mg - kv2)

Then, (mv/(mg - kv2))dv = dx

∫(mv/mg - mv/kv2) dv = ∫dx

Just need a little help integrating if I'm on the right track I guess....

4. Feb 15, 2012

### LawrenceC

Acceleration is not the derivative of velocity with respect to distance; it is the derivative with respect to time.

5. Feb 15, 2012

### jfrank1034

Great....just realized that

6. Feb 15, 2012

### jfrank1034

If I change a to d2s/dt2 is this the right path?

7. Feb 15, 2012

### HallsofIvy

Since your force is given in terms of v, it would be better to write the acceleration as dv/dt:
$$m\frac{dv}{dt}= mg- kv^2$$
and there is no problem with "separating variables"- you have
$$\frac{mdv}{mg- kv^2}= dt$$

8. Feb 15, 2012

### jfrank1034

But I need the velocity in terms of the distance fallen, x. Integrating the right side, ∫dt , would just give me 't' am I right?

9. Feb 15, 2012

### jfrank1034

Was I not correct in saying that a = v (dv/ds) ??

10. Feb 15, 2012

### jfrank1034

In my book it says v dv = a ds --> a = v(dv/ds).....

so I have vdv/(-g + kv2) = dx

∫vdv/(-g + kv2) [from 0 to v] = ∫dx [from 0 to x]

(1/2k) ln(-g+kv2) [from 0 to v] = x

--> (1/2k) ln(g-kv2/g) = x

Solving for v and skipping a few steps I got v = √(g/k(1-e2kx))

11. Feb 15, 2012

### LawrenceC

Your units do not show velocity. You need a mass in there.

12. Feb 15, 2012

### jfrank1034

You're right....should start as dx = mvdv/(-mg + kv2)

13. Feb 15, 2012

### LawrenceC

Whenever you derive something, it is always prudent to check your units in the final product to see that they are what is intended. Secondly, it is a good idea to explore the limits of your formula to see if it makes sense at its limits.