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?

Expert's answer

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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