Astone is dropped into a well. the sound of the splash is heard 3.5 seconds later. what is the depth of the well? take the speed of sound in air to be 343 m/s.

The depth of the well, s = 54.66 m

Given:

time, t = 3.5 s

speed of sound in air, v = 343 m/s

Solution:

By using second equation of motion for the distance traveled by the stone when dropped into a well:

Since, the stone is dropped, its initial velocity, u = 0 m/s

and acceleration is due to gravity only, the above eqn can be written as:

                    (1)

Now, when the sound inside the well travels back, the distance covered,s is given by:

                                             (2)

Now, total time taken by the sound to travel:

t = t' + t''

t'' = 3.5 - t'                                                                        (3)

Using eqn (2) and (3):

s = 343(3.5 - t')                                                                 (4)

from eqn (1) and (4):

Solving the above quadratic eqn:

t' = 3.34 s

Now, substituting t' = 3.34 s in eqn (2)

s = 54.66 m

your answer is, orbit around the nucleus