1. Jun 12, 2013

### Zondrina

1. The problem statement, all variables and given/known data

A boat is floating at rest in dense fog near a large cliff. The captain sounds a horn at water level and the sound travels through the salt water and the air (340 m/s) simultaneously.

The echo in the water takes 0.4s to return. How much additional time will it take for the echo in the air to return?

2. Relevant equations

$v = \frac{Δd}{Δt}$

3. The attempt at a solution

So my take is that $Δt = 0.4s$ for the water echo and a quick look on wiki told me the speed of sound under salt water is $v = 1560 m/s$.

This yields $Δd = vΔt = 624m$.

So the salt water echo takes 0.4 seconds to travel a total distance of 624 m at a rate of 1560 m/s.

Now I'm asked to find how much additional time will it take for the echo in the air to return? So I'm being asked to find $Δt$ for the sound in the air.

I'm given that $v = 340 m/s$ in the air ( By the question ) so the pieces look like they're going to fall together.

I just need to confirm one thing, would i still use $Δd = 624m$ in this next calculation? If so why? If not, why not?

EDIT : I forgot to mention the reason I'm doing this is because I know when I subtract the time it took for the air echo to come back by the time it took for the salt water echo to come back I will get the difference the question is asking for.

EDIT 2 : Nevermind I was reading this question incorrectly. I got my answer.

Last edited: Jun 12, 2013
2. Jan 15, 2017

### Redbelly98

Staff Emeritus
Yes. This is simply the distance that the boat is from the cliff.