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I'm trying to figure out how much distance does a ball of balsa wood covers until reaching terminal velocity, being released from a bottom of a pool.

To my understanding, I need to first figure out the terminal velocity. For that, I take:

$$F_\text{drag} + mg = F_\text{buoyancy}$$

$$F_\text{drag} = C_d A \frac{\rho_m V_t^2}{2}$$ $$F_\text{buoyancy} = V \rho_m g$$ $$mg = V \rho_s g$$

$\rho_m$ is the liquid density, and $\rho_s$ is the ball. $C_d$ is the drag coefficient. A the surface of the ball, $V$ the volume of the ball.

From this equation I reached: $$v_\text{terminal}=\sqrt{\frac{2g}{\gamma}\left(1-\frac{\rho_s}{\rho_m}\right)}$$

$\gamma$ being defined as shape coefficient: $$ \gamma = C_d A / V$$

That looks good, and gives ok results. But when I try to go ahead and find the time / distance I get weird data. I start from this equation: $$m\frac{dv}{dt}=F_b-F_d-mg$$ Solving: $$\frac{dv}{dt}=g\left(\frac{\rho_m}{\rho_s}-1\right) - 0.5\frac{\rho_m}{\rho_s}\gamma v^2$$ with the integral border $v_t$ I found couple of lines before, I get an arctanh of 1, meaning infinity.

I checked the integral, can't find anything wrong. What am I missing here?

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  • $\begingroup$ add the tag of mathematical physics by which you can get an answer from a calculus expert $\endgroup$ – agha rehan abbas Jun 28 '14 at 17:15
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    $\begingroup$ @agharehanabbas No, mathematical-physics is not appropriate for this question. (And adding it would make no difference as to whether you get an answer from a calculus expert) $\endgroup$ – David Z Jun 28 '14 at 17:37
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You're not missing anything. When an object moves through a uniform fluid under the influence of a constant force and drag, the closer it gets to terminal speed, the less the net force on it is and the less it will accelerate. So it only approaches terminal speed asymptotically, never actually reaching it.

As you know, the equation works out to

$$v = v_T\tanh\biggl(\frac{t}{t_0}\biggr)$$

and if you try to find the time at which $v = v_T$ (terminal speed), you do indeed get $\tanh^{-1}(1)$ which is $\infty$. But for any given precision $\delta v$, you can find a time after which $\lvert v - v_T\rvert \leq \delta v$, so given the precision of your measurements, you can tell when the falling object's speed will be indistinguishable from terminal speed.

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  • $\begingroup$ Hmmmm... Sounds about right. I'll try that, thanks $\endgroup$ – Nitay Jun 28 '14 at 20:13
  • $\begingroup$ Another question - t0 doesn't seem to have any physical significance. Does it? I mean, tanh(1) is just a number, I didn't notice anything special $\endgroup$ – Nitay Jun 28 '14 at 21:43
  • $\begingroup$ No, it doesn't really have any physical significance, and that's fine. Though if you work through the math, you find that $t_0 = v_T/g$, so you could think of it as the time it would take you to reach terminal speed without air resistance. (Of course, without air resistance, there wouldn't actually be a terminal speed - that is, the speed would increase without bound, it wouldn't asymptote to any value.) $\endgroup$ – David Z Jun 29 '14 at 3:02

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