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?