Physics example

A speleologist lets fall a stone into a hole and at the beginning it was firm. The sound speed in the air is 343m/s and the speleologist hears the sound of the stone when it touches the hole's bottom 1,5s after he lets it fall.How deep is the hole?

Known data:

Sound speed: vs=343m/sTotal time: t=1.5sSpeed at the beginning: v0=0m/s

Solution:t=t1+t2 where t1=time the stone needs to reach the groundt2=time the sound needs to go back to the speleologist t2=1.5 s – t1.The stone moves with uniformly accelerated rectilinear motion, following the law:

002

2

1stvgts ++= where s0=0m, v0=0m/s and g=9,8m/s2 2

12/9.4 tsms =

Sound propagates with uniform rectilinear motion following the law: 0stvs s += where s0=0m, vs=343m/s stms /343= 2.The distance covered by the sound and the stone are exactly the same, so 2

21

2 /343/9.4 stmtsm = 0/343/9.4 2

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2 =− stmtsm , but t2=1.5s – t1 4.9m/s2 (t1)2 – 343m/s (1.5s – t1)=0 4.9m/s2 (t1)2 + 343m/s t1–514.5m=0 Solving the second degree equation, we obtain two solutions, one is <0 and it is not acceptable (we can’t go back on the time line!), the other one is t1=1.47s t2=1.5 s – 1.47s=0.03s (time needed to the sound to go from the bottom of the hole to the ear of the speleologist) substituting this value into the sound propagation law, we obtain: s=343m/s ∙ 0.03s=10.58m that is the deep of the hole.

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