$$s = ut + \frac{1}{2}at^2$$

where:

* s is the distance the ball travels (12.0 m)

* u is the initial velocity of the ball (0 m/s)

* a is the acceleration due to gravity (-9.8 m/s^2)

* t is the time the ball takes to travel the distance (what we're solving for)

Substituting the given values into the equation, we get:

$$12.0 = 0t + \frac{1}{2}(-9.8)t^2$$

$$12.0 = -4.9t^2$$

Solving for t, we get:

$$t^2 = \frac{12.0}{4.9} = 2.45$$

$$t = \sqrt{2.45} = 1.56\text{ s}$$

Therefore, the ball takes 1.56 seconds to reach the ground.