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 of determining the total time after which the sound of the splash is heard when a ball is dropped into a well of depth \( h \), we can break the problem into two parts: the time taken for the ball to fall to the water surface and the time taken for the sound to travel back up to the top of the well.

### Step-by-Step Solution:

1. **Calculate the time taken for the ball to fall (T1)**:
The ball is falling freely under the influence of gravity. The distance fallen is \( h \), and the initial velocity \( u = 0 \). We can use the equation of motion:
\[ h = \frac{1}{2} g T_1^2 \]
Rearranging this gives:
\[ T_1^2 = \frac{2h}{g} \]
Taking the square root:
\[ T_1 = \sqrt{\frac{2h}{g}} \]

2. **Calculate the time taken for the sound to travel back up (T2)**:
Once the ball hits the water, the sound of the splash travels back to the top of the well. The speed of sound in air is \( c \), and it travels a distance \( h \). The time taken for the sound to travel this distance is given by:
\[ T_2 = \frac{h}{c} \]

3. **Calculate the total time (T)**:
The total time \( T \) after which the sound of 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 up:
\[ T = T_1 + T_2 \]
Substituting the values we found:
\[ T = \sqrt{\frac{2h}{g}} + \frac{h}{c} \]

### Final Answer:
The total time after which the sound of the splash is heard is:
\[ T = \sqrt{\frac{2h}{g}} + \frac{h}{c} \]