A stone is dropped into a well and its splash is heard at the mouth of the well after an interval of 1.45 s.Find the depth of the well. Given that velocity of sound in air at room temperature is equal to `332 ms^(-1)`.

Answer

To find the depth of the well, we can break down the problem into steps.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A stone is dropped into a well, and the splash sound is heard after 1.45 seconds.
- We need to find the depth of the well (denoted as \( h \)).
- The velocity of sound in air is given as \( 332 \, \text{m/s} \).

2. **Setting Up the Time Intervals**:
- Let \( T_1 \) be the time taken for the stone to fall to the bottom of the well.
- Let \( T_2 \) be the time taken for the sound to travel back up to the top of the well.
- The total time is given by:
\[ T_1 + T_2 = 1.45 \, \text{s} \quad \text{(Equation 1)} \]

3. **Calculating \( T_1 \)**:
- The time \( T_1 \) for the stone to fall can be calculated using the formula for free fall:
\[ T_1 = \sqrt{\frac{2h}{g}} \]
- Here, \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \).

4. **Calculating \( T_2 \)**:
- The time \( T_2 \) for the sound to travel back up can be calculated using:
\[ T_2 = \frac{h}{v} \]
- Where \( v \) is the speed of sound, \( 332 \, \text{m/s} \).

5. **Substituting \( T_1 \) and \( T_2 \) into Equation 1**:
- Substitute the expressions for \( T_1 \) and \( T_2 \) into Equation 1:
\[ \sqrt{\frac{2h}{g}} + \frac{h}{332} = 1.45 \]

6. **Solving for \( h \)**:
- To solve for \( h \), we can square both sides of the equation to eliminate the square root:
\[ \left(\sqrt{\frac{2h}{g}} + \frac{h}{332}\right)^2 = (1.45)^2 \]
- This will yield a quadratic equation in terms of \( h \).

7. **Calculating the Depth**:
- After solving the quadratic equation, we find that \( h \) is approximately \( 9.9 \, \text{m} \).

### Final Answer:
The depth of the well is approximately **9.9 meters**.