Two teams of movers are lowering a piano from the window of a $10$ floor apartment building. The rope breaks
when the piano is $30$ meters above the ground. The movers on the ground, alerted by the shouts of the movers
above, first notice the piano when it is $14$ meters above the ground. How long do they have to get out of the way
before the piano hits the ground?

(A) $\text{0.66 sec}$
(B) $\text{0.78 sec}$
(C) $\text{1.67 sec}$
(D) $\text{1.79 sec}$
(E) $\text{2.45 sec}$
1. **Determine the time it takes for the piano to fall from 30 meters to the ground:**

   The equation for the distance fallen under gravity is given by:
   \[
   h = \frac{1}{2}gt^2
   \]
   where \( h \) is the height, \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)), and \( t \) is the time in seconds.

   For \( h = 30 \) meters:
   \[
   30 = \frac{1}{2} \cdot 9.8 \cdot t^2
   \]
   Solving for \( t \):
   \[
   30 = 4.9t^2
   \]
   \[
   t^2 = \frac{30}{4.9}
   \]
   \[
   t = \sqrt{\frac{30}{4.9}}
   \]
   \[
   t \approx \sqrt{6.1224}
   \]
   \[
   t \approx 2.47 \, \text{seconds}
   \]

2. **Determine the time it takes for the piano to fall from 14 meters to the ground:**

   For \( h = 14 \) meters:
   \[
   14 = \frac{1}{2} \cdot 9.8 \cdot t^2
   \]
   Solving for \( t \):
   \[
   14 = 4.9t^2
   \]
   \[
   t^2 = \frac{14}{4.9}
   \]
   \[
   t = \sqrt{\frac{14}{4.9}}
   \]
   \[
   t \approx \sqrt{2.8571}
   \]
   \[
   t \approx 1.69 \, \text{seconds}
   \]

3. **Calculate the time the movers have to get out of the way:**

   The time difference between the piano being at 30 meters and 14 meters is:
   \[
   t_{\text{difference}} = \sqrt{\frac{30}{4.9}} - \sqrt{\frac{14}{4.9}}
   \]
   \[
   t_{\text{difference}} \approx 2.47 - 1.69
   \]
   \[
   t_{\text{difference}} \approx 0.78 \, \text{seconds}
   \]

The final answer is \(\boxed{0.78 \, \text{seconds}}\)