Set up the equation of motion.

The forces acting on the ball are gravity downwards ($mg$) and resistive force upwards ($mg \frac{v^2}{c^2}$).
Using Newton's Second Law, $F_{net} = ma$:

$$m\frac{dv}{dt} = mg - mg\frac{v^2}{c^2}$$

Divide by $m$:

$$\frac{dv}{dt} = g\left(1 - \frac{v^2}{c^2}\right)$$

Rearrange the equation to separate $v$ and $t$:

$$\frac{dv}{1 - \frac{v^2}{c^2}} = g\,dt$$ $$\frac{c^2\,dv}{c^2 - v^2} = g\,dt$$

Integrate both sides. The left side is a standard integral of the form $\int \frac{1}{a^2 - x^2} dx = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right|$. Here, $a=c$ and $x=v$.

$$\int \frac{c^2}{c^2 - v^2}\,dv = \int g\,dt$$ $$c^2 \cdot \frac{1}{2c} \ln \left| \frac{c+v}{c-v} \right| = gt + K$$ $$\frac{c}{2} \ln \left| \frac{c+v}{c-v} \right| = gt + K$$

The ball is dropped, so at $t=0$, $v=0$.

$$\frac{c}{2} \ln \left| \frac{c+0}{c-0} \right| = g(0) + K$$ $$\frac{c}{2} \ln(1) = K$$ $$K = 0$$

Substitute $K=0$ back into the equation:

$$\frac{c}{2} \ln \left( \frac{c+v}{c-v} \right) = gt$$

(We can remove the absolute value since $v < c$ for the ball to be accelerating, so $c+v$ and $c-v$ are positive.)

$$t = \frac{c}{2g} \ln \left( \frac{c+v}{c-v} \right)$$

Substitute $v = \frac{c}{2}$ into the expression for $t$:

$$t = \frac{c}{2g} \ln \left( \frac{c + \frac{c}{2}}{c - \frac{c}{2}} \right)$$ $$t = \frac{c}{2g} \ln \left( \frac{\frac{3c}{2}}{\frac{c}{2}} \right)$$ $$t = \frac{c}{2g} \ln(3)$$

The time taken for the ball to attain a speed of $\frac{c}{2}$ is $\boxed{\frac{c}{2g} \ln(3)}$.