Find distance using speed of sound

  • Thread starter L1988
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  • #1
L1988
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Two questions:
1. A sailor strikes the side of his ship just below the waterline. He hears the echo of the sound reflected from the ocean floor directly below 2.02 s later. How deep is the ocean at this point?

I assumed this used distance = speed x time. I knew that speed must equal the speed of sound in water, which I think is 1500 m/s (pretty sure, but not 100%) and I also divided my final distance by 2 so that I only got the distance of depth one way. However, the computer (this is online homework) said I was wrong, so I'm not sure where I went wrong. Any help would probably help me with the next problem too.


2.
A rescue plane flies horizontally at a constant speed searching for a disabled boat. When the plane is directly above the boat, the boat's crew blows a loud horn. By the time the plane's sound detector perceives the horn's sound, the plane has traveled a distance equal to one half its altitude above the ocean. If it takes the sound 2.17 s to reach the plane, determine the altitude of the plane. Take the speed of sound to be 350 m/s.

Just like last time, I assumed that I should use

distance= speed of sound x time

with a new speed (350 m/s), new time, and this time distance doesn't have to be divided by 2. However, that got me the wrong answer.

Then I tried drawing a triangle, (since the plane has moved) and called my final distance r^2 and my x^2 = one half its altitude, so all that was left was y^2. This also was incorrect.

Where am I making a mistake?
 

Answers and Replies

  • #2
DaveC426913
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Well, what WERE your answers? How can we see where you went wrong if you don't show us what you got?
 
  • #3
L1988
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sorry, I am new to this forum.

for the first one I got 1515m
343m/s x 2.02 s = 3030 m
3030m /2 = 1515 m
 
  • #4
L1988
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second question:
Note: problem says : Take the speed of sound to be 350 m/s.

attempt 1:
distance = speed x time
d = 350 x 2.17 s = 759.5 m

attempt 2: using a triangle and pythagorean's theorem

759.5 = r^2
x^2 = "distance equal to one half its altitude above the ocean"
x^2= (sqrt (759.5) /2)^2
Y^2 = r^2-x^2
y^2= 759.5-189.9
y^2= 23.9m
 
  • #5
hage567
Homework Helper
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343m/s x 2.02 s = 3030 m

Check your arithmetic here.
 
  • #6
L1988
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sorry i meant speed of sound
1500 x 2.02
 
  • #7
hage567
Homework Helper
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second question:
Note: problem says : Take the speed of sound to be 350 m/s.

attempt 1:
distance = speed x time
d = 350 x 2.17 s = 759.5 m

attempt 2: using a triangle and pythagorean's theorem

759.5 = r^2
x^2 = "distance equal to one half its altitude above the ocean"
x^2= (sqrt (759.5) /2)^2
Y^2 = r^2-x^2
y^2= 759.5-189.9
y^2= 23.9m

How did you get the 189.9? If you don't know the altitude, how can you know half the altitude?? You can express the two sides of the triangle in terms of the altitude, h (what you are trying to find). Yes, 759.5 m is the length of the hypotenuse of your triangle, but make sure you are using the Pythagorean theorem correctly (be careful of your algebra).
 
  • #8
L1988
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ok:

759.5 = (h/2) ^2 + h^2
759.5 = (h^2 / 4) + (h)^2
759.5 = (h)^2 (1+1/4)
759.5 = h^2 (1.25)
759.5 / (1.25) = h^2
h^2 = 607.6
h = 24.6

Still incorrect according to the computer
 
  • #9
hage567
Homework Helper
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Don't forget to square the 759.5 meters.
 
  • #10
L1988
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awesome! I got it,

final answer = 679.3

Thanks for your help!

I don't think I can use this method for the boat question, right? Do you have any suggestions for that?
 
  • #11
DaveC426913
Gold Member
21,181
4,612
ok:

759.5 = (h/2) ^2 + h^2
759.5 = (h^2 / 4) + (h)^2
759.5 = (h)^2 (1+1/4)
759.5 = h^2 (1.25)
759.5 / (1.25) = h^2
h^2 = 607.6
h = 24.6

Still incorrect according to the computer

BTW, one of the things you hopefully are learning/practicing is how to check your answers intuitively. When you found the answer, did you stop and ask yourself "does it make sense that the airplane was flying less than 25 metres above the ocean?"
 
  • #12
L1988
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I still don't understand why

(1500 m/s) x (2.02s) x (.5) = 1515 m

is incorrect for the boat problem
 
  • #13
DaveC426913
Gold Member
21,181
4,612
I still don't understand why

(1500 m/s) x (2.02s) x (.5) = 1515 m

is incorrect for the boat problem

Frankly, I don't understand how an online service deals with margins of error. What if the correct answer is 1514?
 

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