Question

a stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. determine the depth of the well.

Answer 1

\[d=ut+1/2at ^{2}\]let,
d =?
a = 9.8 m/s (aproximately the acceleration of the earth,
you can use 10)
u=0
Then just solve for d.
\[d = 0 \times 3.41 + 1/2 \times 9.8 \times 3.14^{2}\]

Answer 2

The answer needs to take in to account the travel time of sound back up to the top of the well. If you let \[t_1\] be the time taken for the rock to hit the water and \[t_2\] be the time taken for the sound of the splash to reach the top of the well. Then \[t_1+t_2 = 3.41s\] This means that \[t_2 = 3.41-t_1\] Your textbook should provide a constant for the speed of sound (with the assumption your at 20 degrees C or something similar)
If we assume the speed of sound is 343m/s and let the height of the cliff
be represented by d then \[t_2 = d/343\] so \[d = 343t_2\] This means that \[d = 343(3.41-t_1)\] You can then substitute this into
\[d = (1/2)a{t_1}^2\] So
\[343(3.41-t_1) = (1/2)(9.8){t_1}^2\] Then all you have to do is solve the quadratic equation for \[t_1\] and plug the result into the original equation for the height of the cliff i.e. \[d=(1/2)(9.8){t_1}^2\]