A ball is dropped into a well in which the water level is at a depth `h` below the top. If the speed of sound id `C`, then the time after which the splash is heard will be give by.

Options

A. `h[sqrt((2)/(gh))+(1)/(c)]`
B. `h[sqrt((2)/(gh))+(1)/(c)]`
C. `h[(2)/(g)+(1)/(c)]`
D. `h[(2)/(g)+(1)/(c)]`

Explanation

To solve the problem of finding the total time after which the splash is heard when a ball is dropped into a well, we will break the solution into clear steps.

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Let `h` be the depth of the well.
- Let `C` be the speed of sound.
- The initial velocity (`u`) of the ball when dropped is 0.

2. **Calculate the Time for the Ball to Fall**:
- The equation of motion for the ball falling under gravity is given by:
\[ h = ut + \frac{1}{2}gt^2 \]
- Since the ball is dropped, `u = 0`, so the equation simplifies to:
\[ h = \frac{1}{2}gt_1^2 \]
- Rearranging gives:
\[ t_1^2 = \frac{2h}{g} \]
- Taking the square root:
\[ t_1 = \sqrt{\frac{2h}{g}} \]

3. **Calculate the Time for Sound to Travel Back**:
- After the ball hits the water, the sound travels back up to the top of the well. The time taken for sound to travel a distance `h` at speed `C` is given by:
\[ t_2 = \frac{h}{C} \]

4. **Calculate the Total Time**:
- The total time `T` after which the splash is heard is the sum of the time taken for the ball to fall and the time taken for the sound to travel back:
\[ T = t_1 + t_2 \]
- Substituting the values of `t_1` and `t_2`:
\[ T = \sqrt{\frac{2h}{g}} + \frac{h}{C} \]

5. **Final Expression**:
- Thus, the total time after which the splash is heard is:
\[ T = \sqrt{\frac{2h}{g}} + \frac{h}{C} \]