Question

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.90s apart. The speed of sound in air is343 m/s, and in concrete is 3000 m/s.

How far away did the impact occur?

Answer 1

We apply d = vt for the two sounds, and the distance is the same for both...

Thus vt = vt

(3000)(t) = (343)(t + .9)

3000t = 343t + 308.7

2657t = 308.7

t = .116 sec

d = vt

d = (3000)(.116)

d = 348.6 m

Answer 2

Considering time taken by the light is almost nill

lets assume the distance is x

==> Time taken by sound in air(t1) = x/v = x/343

==> Time taken by sound in concrete(t2) = x/v = x/3000

It is given that t1-t2=0.55 s

==> x*(1/343 - 1/3000) = 0.55

==> x*0.00258 = 0.55

==> x = 213 m