How Do Echo Timings Help Calculate Cliff Distances?

[the_storm]

Homework Statement

A cowboy stands on horizontal ground between two paral-
lel vertical cliffs. He is not midway between the cliffs. He
fires a shot and hears its echoes. The second echo arrives
1.92 s after the first and 1.47 s before the third. Consider
only the sound traveling parallel to the ground and reflect-
ing from the cliffs. Take the speed of sound as 340 m/s.

Homework Equations

(a) What is the distance between the cliffs? (b) What If? If
he can hear a fourth echo, how long after the third echo
does it arrive?

The Attempt at a Solution

I aid the first echo happens when the sound travels a distance = 2X₁ where X₁ is the distance between the near cliff
and the second echo happens at 2X₂ and X₂ is the distance between the source of the sound and the far cliff
So
V=\frac{X}{T} >> V=\frac{ 2X₁ }{ T } >>> eq1
V=\frac{ 2X₂ }{ 1.92T } >>> eq2
divide 1 over2
so I get X₂ =1.9 X₁
Is it correct ? if that is correct so what about the third echo when it happens ?

 

[Onamor]

I think this one needs a few reads.
When it says "The second echo arrives ... 1.47 s before the third." Doesn't that then make 1.47 the time for the first echo to arrive?

 

[Gear.0]

Indeed. Also, which sound wave will be the third echo?

Of course neither of those things are required to know, it should all be handled automagically by the equations.

But, why did you use 1.92T for the time of the second echo? I think it should be the sum not the product because the second echo arrives 1.92 seconds after the first echo (which took time T) so the second echo took (1.92 + T) seconds.

It will be similar for the third echo.
The thing that you have to think about a bit for the third echo, is what distance did it travel?

 

[Onamor]

automagically

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