A stone is dropped into a well in which the level of water is H metre below the top of the well. If v is velocity splash is heard will be given by

Options

A. `(2H)/(V)`
B. `sqrt((2H)/(g))+(H)/(V)`
C. `sqrt((H)/(2g))+(2H)/(V)`
D. `sqrt((2H)/(V))+(H)/(g)`

Answer

To solve the problem of finding the total time taken for the splash sound to be heard after a stone is dropped into a well, we can break it down into a few steps.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A stone is dropped from the top of a well that is H meters deep.
- We need to find the total time taken for the splash sound to reach the observer after the stone hits the water.

2. **Time for the Stone to Fall (T1)**:
- The stone is dropped from rest, so its initial velocity (u) is 0.
- We can use the second equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
Here, S = H (the depth of the well), a = g (acceleration due to gravity), and u = 0.
- Thus, the equation simplifies to:
\[ H = \frac{1}{2} g T_1^2 \]
- Rearranging gives:
\[ T_1^2 = \frac{2H}{g} \]
- Taking the square root:
\[ T_1 = \sqrt{\frac{2H}{g}} \]

3. **Time for Sound to Travel Back Up (T2)**:
- After the stone hits the water, the sound travels back up to the observer.
- The distance the sound travels is also H meters, and the speed of sound is v.
- The time taken for sound to travel back up is given by:
\[ T_2 = \frac{H}{v} \]

4. **Total Time (T)**:
- The total time taken for the splash sound to be heard is the sum of T1 and T2:
\[ T = T_1 + T_2 \]
- Substituting the values we found:
\[ T = \sqrt{\frac{2H}{g}} + \frac{H}{v} \]

5. **Final Expression**:
- Therefore, the total time taken for the splash sound to be heard is:
\[ T = \sqrt{\frac{2H}{g}} + \frac{H}{v} \]