1. 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/s
Total time: t=1.5s ?
Speed at the beginning: v0=0m/s
Solution:
t=t1+t2 where t1=time the stone needs to reach the ground
t2=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:
1 2
s= gt + v 0 t + s 0 where s0=0m, v0=0m/s and g=9,8m/s2  s = 4.9m / s 2 t12
2
Sound propagates with uniform rectilinear motion following the law: s = v s t + s 0 where s0=0m,
vs=343m/s  s = 343m / st 2.
The distance covered by the sound and the stone are exactly the same, so 4.9m / s 2 t12 = 343m / st 2
 4.9m / s t1 − 343m / st 2 = 0 , but t2=1.5s – t1  4.9m/s2 (t1)2 – 343m/s (1.5s – t1)=0 
2 2
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.