Water in a well is at a depth h. A ball is released from ground so that it falls into the well. The speed of sound in air is c. The time after which sound of splash is heard is :

Options

A. `sqrt (( 2h )/( g ) ) + (2h )/( c) `
B. ` sqrt ((2h )/(g) )+ (h)/©`
C. ` sqrt ((2h )/( g ) ) + (h )/(2c) `
D. ` sqrt((h)/(g) ) +(h)/© `

Explanation

To solve the problem, we need to determine the total time taken for a ball to fall into a well and for the sound of the splash to travel back up to the point of release.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Depth of the well = \( h \)
- Acceleration due to gravity = \( g \)
- Speed of sound in air = \( c \)

2. **Calculate the Time for the Ball to Fall:**
- When the ball is released from the ground, it falls freely under the influence of gravity.
- The initial velocity \( u = 0 \).
- The distance fallen \( s = h \).
- Using the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the known values:
\[ h = 0 \cdot t_1 + \frac{1}{2} g t_1^2 \]
This 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. **Calculate the Time for Sound to Travel Back:**
- After the ball hits the water, the sound travels back up to the point of release.
- The distance the sound travels is also \( h \).
- The time taken for sound to travel this distance is given by:
\[ t_2 = \frac{h}{c} \]

4. **Calculate the Total Time:**
- The total time \( T \) 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 expressions for \( t_1 \) and \( t_2 \):
\[ T = \sqrt{\frac{2h}{g}} + \frac{h}{c} \]

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