Let the time the stone spent falling in the air be \( t_a \) seconds and the time it spent falling in the water be \( t_w \) seconds. According to the problem, the total time of the fall is:
\[ t_a + t_w = 12 \quad \text{(1)} \]
The total distance fallen is \( 127 \) meters, which can be divided into the distance fallen in the air and the distance fallen in the water.
The distance fallen in the air can be calculated using the formula:
\[ d_a = v_a \cdot t_a = 16 \cdot t_a \quad \text{(2)} \]
The distance fallen in the water can be expressed as:
\[ d_w = v_w \cdot t_w = 3 \cdot t_w \quad \text{(3)} \]
The total distance fallen is the sum of the distances in air and water:
\[ d_a + d_w = 127 \quad \text{(4)} \]
Substituting equations (2) and (3) into equation (4):
\[ 16 \cdot t_a + 3 \cdot t_w = 127 \quad \text{(5)} \]
Now we have a system of equations (1) and (5):
- \( t_a + t_w = 12 \)
- \( 16 \cdot t_a + 3 \cdot t_w = 127 \)
We can solve for \( t_w \) in terms of \( t_a \) from equation (1):
\[ t_w = 12 - t_a \]
Substituting this expression for \( t_w \) into equation (5):
\[ 16 \cdot t_a + 3 \cdot (12 - t_a) = 127 \]
Expanding this gives:
\[ 16 \cdot t_a + 36 - 3 \cdot t_a = 127 \]
Combining like terms results in:
\[ 13 \cdot t_a + 36 = 127 \]
Subtract 36 from both sides:
\[ 13 \cdot t_a = 91 \]
Now, divide by 13:
\[ t_a = \frac{91}{13} \approx 7 \quad \text{(6.999)} \text{ seconds} \]
Now we calculate \( t_w \):
\[ t_w = 12 - t_a \approx 12 - 7 = 5 \text{ seconds} \]
Thus, the stone fell in the air for approximately \( 7 \) seconds and fell in the water for \( 5 \) seconds.
To conclude, the answers are:
- The stone fell in the air for \( 7 \) seconds.
- The stone fell in the water for \( 5 \) seconds.