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

Question

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?

Answer 1

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.