# A stone is dropped into a well. The sound of the splash is heard 3.5 seconds later. What is the depth of the well? Take the speed of sound in air to be 343 m/s.

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Answer:The depth of the well, s = 54.66 mGiven:time, t = 3.5 sspeed of sound in air, v = 343 m/sSolution:By using second equation of motion for the distance traveled by the stone when dropped into a well:[tex]s = ut +\frac{1}{2}at^{2}[/tex]Since, the stone is dropped, its initial velocity, u = 0 m/sand acceleration is due to gravity only, the above eqn can be written as:[tex]s = \frac{1}{2}gt'^{2}[/tex][tex]s = \frac{1}{2}9.8t^{2} = 4.9t'^{2}[/tex] (1)Now, when the sound inside the well travels back, the distance covered,s is given by:[tex]s = v\times t''[/tex][tex]s = 343\times t''[/tex] (2)Now, total time taken by the sound to travel:t = t' + t'' t'' = 3.5 - t' (3)Using eqn (2) and (3):s = 343(3.5 - t') (4)from eqn (1) and (4):[tex]4.9t'^{2} = 343(3.5 - t')[/tex] [tex]4.9t'^{2} + 343t' - 1200.5 = 0[/tex] Solving the above quadratic eqn:t' = 3.34 s Now, substituting t' = 3.34 s in eqn (2)s = 54.66 m