# Terminal Velocity of buoyant object

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?

• add the tag of mathematical physics by which you can get an answer from a calculus expert – agha rehan abbas Jun 28 '14 at 17:15
• @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) – David Z Jun 28 '14 at 17:37

$$v = v_T\tanh\biggl(\frac{t}{t_0}\biggr)$$
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.
• 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.) – David Z Jun 29 '14 at 3:02