Determine the depth of a well in which a stone is dropped, and where you can hear the impact on the water after 1.5 s, considering that the velocity of sound is 340 m/s.



t = 1.5 s
vsound = 340 m/s
g = 9.8 m/s2
H = ?


Let's consider the exercise in two parts:

  • Part 1. The stone descends to impact the water at the bottom of the well
  • Part 2. The sound moves up from the water to the opening of the well

If we call the time the stone takes to get to the water t1 and t2 the time the sound takes to get back to the opening of the well, we get that:


We will study each part separately:

Part 1

Applying the equation of position of the fall free motion and knowing that at t1, the position of the stone is y = 0 m:

y=H-12·g·t2 0=H-0.5·9.8·t12 H=4.9·t12

Part 2

Sound rises at constant speed, and we assume that it does it in a straight line. Therefore:

x=v·t H=vsonido·t2 H=340·t2

Since H is the same in both equations and we know that t=t1+t2:

4.9·t12=340·t21.5=t1+t2 4.9·t12-340·t2=0t2=1.5-t1 4.9·t12-340·1.5·t1=0 4.9·t12-340·t1-510=0 t1=-340±3402-1·4.9·(-510)2·4.9 t1=1.47 s

Substituting t1 in the equation of part 1:

H=4.9·1.472 H=10.59 m

Formulas worksheet

These are the main formulas that you must know to solve this exercise. If you are not clear about their meaning, we recommend you to check the theory in the corresponding sections. Furthermore, you will find in them, under the Formulas tab, the codes that will allow you to integrate these equations in external programs like Word or Mathematica.

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