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# Answer to Question #123862 in Mechanics | Relativity for ravi tetarwal

Question #123862
You drop a rock into a well and count the seconds before you hear a splash. If you counted 10 seconds before you heard the splash, how deep is the well?
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Expert's answer
2020-06-25T09:35:05-0400

Let "h" be the depth of the well. 10 seconds is the time for rock to fall to water and for sound to travel to us.

"h = \\dfrac{gt_1^2}{2}, \\;\\;\\;" "h = c_st_2, \\;\\;\\; t_1+t_2=10\\,\\mathrm{s}."

Therefore, "gt_1^2=2c_st_2, \\;\\; t_2=10-t_1." So "gt_1^2=2c_s(10-t_1)." Solving this equation, we get "t_1=8.8\\,\\mathrm{s}, \\; t_2 = 1.2\\,\\mathrm{s}, h \\approx 396\\,\\mathrm{m}."

We may notice that the time interval is quite large, so the well has significant depth.

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