# Terminal velocity in finite time

1. Apr 24, 2006

### cscott

$$v^2 = \frac{mg}{k}(1 - e^{-2ky/m})$$

As $t \rightarrow \infty$ and $y \rightarrow \infty$ we see $TV = \sqrt{mg/k}$

And below my book reads: "From actual experience we know that a raindrop reaches its limiting velocity in a finite and not an infinite amount of time. This is because other factors also operate to slow the raindrop's velocity."

What are these factors?

Last edited: Apr 24, 2006
2. Apr 24, 2006

### Cyrus

I don't know what he would mean by this. The drag force is what's slowing down the raindrop, so friction can't be an answer. K depends on the geometry of the raindrop and the fluid. The K should change a small amount as the density of the fluid increases as the raindrop gets lower. But appart from that, I don't know. Maybe Clausius can tell us why.

3. Apr 24, 2006

### dav2008

Maybe the passage means that the fluctuations in the value of k as the object falls change the velocity more than the difference between v and vt

4. Apr 25, 2006

### cscott

I think you're right because before he states some approximations so that air resistance is R = kv.