A 67-kg base runner begins his slide into second base when he is moving at a speed of 3.6 m/s. The coefficient of friction betwe
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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:
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