(a) The distance to the water, when the time taken by sound to travel up the well is neglected is 19.620 m.
(b) When the time taken by sound is also taken into consideration, is 18.5395 m.
(a) When a rock dropped into a well of height h it falls freely under the action of acceleration due to gravity g. If the time taken by the sound to reach up the well can be neglected, then,
[tex]h=\frac{1}{2} gt^2[/tex]
therefore,
[tex]h=\frac{1}{2} gt^2\\ =\frac{1}{2}(9.81m/s^2)(2.0000s)^2\\ =19.620 m[/tex]
the depth of the well is found to be 19.620 m
(b) The rock takes a time t to reach the water and the sound of splash travels at the speed of sound v upwards and takes a time (2.0000-t)s to reach the top of the well.
Therefore,
[tex]\frac{1}{2} gt^2=v(2.0000-t)[/tex]
Substitute 9.81m/s²for g and 332.00 m/s for v.
[tex]\frac{1}{2} gt^2=v(2.0000-t)\\ 4.905t^2+332.00t-664.00=0[/tex]
Solve the quadratic equation.
[tex]t=\frac{(-332.00)+/-( \sqrt{(332.00)^2+(4(4.905)(664.00)} }{2(4.905)} \\ =\frac{(-332.00)+/-(351.07218)}{9.81}[/tex]
Taking the positive value alone,
[tex]t=1.94415s[/tex]
The depth of water in the well is given by,
[tex]h=\frac{1}{2} gt^2\\ =\frac{1}{2}(9.81 m/s^2)(1.94415s)^2\\ =18.5395 m[/tex]
Thus the depth of the well is found to be 18.5395 m
