Finding Distance from a Crash with Asymmetric Hearing

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A swimmer hears a crash underwater and then again in air after 2 seconds. The speed of sound in water is 1500 m/s, while in air at 20°C, it is approximately 343.8 m/s. The time taken for sound to travel in water and air can be expressed in terms of the distance from the crash. By setting up equations based on the time it takes for sound to travel in both mediums, the swimmer can determine the distance. The discussion emphasizes solving for distance using the relationship between the times and speeds of sound in water and air.
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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...
 
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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...
 
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?
 
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.
 
k thanks ill try that
 

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