Finding out the speed of sound

[Creed]

Homework Statement

Two persons, A and B were interest to find out the speed of the sound in the air. They stand 40 m of a wall and 60 m from each other. The person B hears the sound that A said. And after 1/8 seconds he hears the echo created by the wall. Based on this measures, what is the value these two people obtained for the speed of sound?

Solution: 340 m/s

Relevant Equations

v = d/t

d² = 40² + 60² ---> I

60 = V.t ---> II

40 + d = V.(t + 1/8) ---> III

Thanks in the advance. :)

 

[TSny]

If A wanted to throw a ball such that it went to B by first bouncing off the wall, at what point would the ball need to hit the wall?

 

[Creed]

In the middle of it.

 

[Merlin3189]

I can see a reason for I and II, despite your lack of explanation, but not for III.
Maybe you can mark d on your diagram. I only know where d is from eqn I. I do not know the significance of d.

I wonder what the significance of the grey arrows is? B is looking at A and A is looking at the wall??
I think they are not the paths of the two sound signals.
Perhaps you should show the paths of the two sound signals.

But your approach is right. I think you need to be clearer about what's happening.
Also I can't get 340 m/s by your idea nor mine, so I don't know what they mean by saying it is the solution.

Edit: To get a speed of 340 m/sec when the difference in time is 1/8 sec, they would need a path difference of (1/8) sec x 340 m / sec = 42.5 m

 

[Creed]

Thanks for helping out, Merlin 3189 and Tsny. I really appreciate it.

Well observed.

There's a mistake, the solution is 320 m/s. The grey arrows should have the end curved in both sides, so A is looking at B and B looking at A and A is looking at wall and wall is facing A. I'm sorry for the draw I had to made at the computer with limited resources.

I just solved the exercise.

With different approach using what Tsny asked me. We have to considered the sound wave bouncing back in the middle of the wall. And using the Pythagorean theorem to find out the distance of wave goes.
That is 50 m from A to middle point of the wall plus 50m from middle point of the wall to B, 100m in Total.

So if, first sound wave travels 60 m.
The second sound wave that goes to A to the middle point of the wall also travels 60 m. Therefore, there is only 40 m left to travel.

Therefore:

v = d/t <=> v = 40 / (1/8)<=> v=320 m/s.

Thank you again to both you.

 

[Merlin3189]

I did also wonder why you didn't get an answer from your 3 equations:
I ⇒ d =√5200 = 72.11
III ⇒ 40+72.11 = Vt + 0.125 V
⇒ 112.11 = 60 + 0.125 V
⇒ V = 8 x 52.11 = 417 m/sec