Respuesta :
Answer:
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:
[tex]s = ut +\frac{1}{2}at^{2}[/tex]
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:
[tex]s = \frac{1}{2}gt'^{2}[/tex]
[tex]s = \frac{1}{2}9.8t^{2} = 4.9t'^{2}[/tex] (1)
Now, when the sound inside the well travels back, the distance covered,s is given by:
[tex]s = v\times t''[/tex]
[tex]s = 343\times t''[/tex] (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):
[tex]4.9t'^{2} = 343(3.5 - t')[/tex]
[tex]4.9t'^{2} + 343t' - 1200.5 = 0[/tex]
Solving the above quadratic eqn:
t' = 3.34 s
Now, substituting t' = 3.34 s in eqn (2)
s = 54.66 m
