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 0.50 s apart. The speed of sound in air is 343 m/s, and in concrete is 3000 m/s.

Required:
How far away did the impact occur?

Answer 1

The distance is
`d = 193.6 \ m`

Step-by-step explanation:

From the question we are told that

The time interval between the sounds is k
`t_1 = k + t_2` = 0.50 s

The speed of sound in air is
`v_s = 343 \ m/s`

The speed of sound in the concrete is
`v_c = 3000 \ m/s`

Generally the distance where the collision occurred is mathematically represented as

`d = v * t`

Now from the question we see that d is the same for both sound waves

So

`v_c t = v_s * t_1`

Now

So
`t_1 = k + t`

`v_c t = v_s * (t+ k)`

=>
`3000 t = 343* (t+ 0.50)`

=>
`3000 t = 343* (t+ 0.50)`

=>
`t = 0.0645 \ s`

So

`d = 3000 * 0.0645`

`d = 193.6 \ m`