Question

A stone is dropped into a well in which the level of water is h below the top of the well. If v is speed of sound in the air, the time T after which the splash is heard is given by

a. T = 2h/v
b. T = √2h/g + h/v
c. T = √2h/g + h/2v
d. T = √h/2g + 2h/v

Answer 1

The time T after which the splash is heard when a stone is dropped into a well is given by the sum of the time the stone takes to fall to the water's surface and the time for the sound to travel back up. The correct formula is T = sqrt(2h/g) + h/v, option B is correct.

Step-by-step explanation:

The question asks for the calculation of time T after which the splash is heard when a stone is dropped into a well with water level h below the top, considering v as the speed of sound in the air. To find the correct formula for the time T, we need to consider two separate parts of the motion: (1) the time taken for the stone to fall to the water's surface, and (2) the time taken for the sound to travel back up the well to the listener.

• The time t1 it takes for the stone to fall is found using the equation of motion under gravity t1 = sqrt(2h/g), where g is the acceleration due to gravity.
• The time t2 it takes for the sound to travel back up the well to reach the listener is t2 = h/v, where h is the height and v is the speed of sound.
• The total time T heard after the stone is dropped would be the sum of these two times, T = t1 + t2, which calculates to T = sqrt(2h/g) + h/v.

Looking at the provided options, the answer that matches our calculation is option B: T = sqrt(2h/g) + h/v. This is the correct option.