Finding Distance from a Crash with Asymmetric Hearing

[PiRsq]

Sound Problem...

A swimmer with one ear underwater and one ear above water in air, hears a crash. She hears the crash in the water first where the Veloctiy of the wave in water was 1500m/s. The air temperature is 20°C and she hears the wave again after 2.0 seconds but this time she hears it through the ear that's above the water. Find the distance from the crash...


What I tried was this:

-Vair=332 + 0.59 (20°) = 343.8m/s

-Since she hears the sound in air 2 seconds after it must've traveled at least 343.8*2 = 687.6 m

So is the distance is 687.6 m?


Thx for help...

 

[gnome]

Try again.

Let t = time it took for the sound to arrive in water.

Now, in terms of t, how long did it take in air?

See if you can do anything with that...

 

[PiRsq]

Im sorry I don't understand...If i find the ratio of the time it took for water than air, how will that help me?

 

[FZ+]

Ok, you know the speed of sound in air at 20 degrees, right?

let x be the distance...
x/1500 = t1 (time taken in water)

x/speed of sound in air (20C) = t2

t2 = t1 + 2 (since it is two seconds later)

x/speed of sound in air(20 C) = x/1500 + 2

Now, solve.

 

[PiRsq]

k thanks ill try that