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

Answer:

The distance is $d = 193.6 \ m$

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

     $d = 3000 * 0.0645$

     $d = 193.6 \ m$