How To Find Depth Given TIME ONLY

  • Thread starter misz_sunshine
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In summary: The distance between the ship and the cliff is 440 meters. The sound has to travel440 meters to reach the cliff and back. Given this information, it will take 1.2 seconds for the sound to travel.
  • #1
misz_sunshine
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Homework Statement


The time it takes a high frequency sound to travel from the surface of the lake to the bottom of the lake and back to the surface is 0.1seconds. What is the depth of the lake?


Homework Equations


Speed of sound in water= 1500m/s
Time=0.1s
Depth=?


The Attempt at a Solution


Depth= Speed of sound * time elapsed/2= 1500*0.1/2
= 75m
IS THIS CORRECT? PLEASE HELP?
 
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  • #2
Looks good to me!
 
  • #3
Really? I assumed that 1500 m/s was the speed of sound in water. Didn't know if it was correct since it wasn't given to me in the question... Can u help me with another question please?
 
  • #4
Really? I assumed that 1500 m/s was the speed of sound in water. Didn't know if it was correct since it wasn't given to me in the question... Can u help me with another question please?
 
  • #5
Oh, I assumed you were given the speed of sound in water at 1500 m/s, but it still looks good, since Wikipedia says the speed of sound in water is 1484 m/s (http://en.wikipedia.org/wiki/Speed_of_sound). What is your other question?
 
  • #6
phyzguy said:
Oh, I assumed you were given the speed of sound in water at 1500 m/s, but it still looks good, since Wikipedia says the speed of sound in water is 1484 m/s (http://en.wikipedia.org/wiki/Speed_of_sound). What is your other question?

Thanks.
The other question is:
A ship is 220metres from a large cliff when it sounds its foghorn. If the echo time is 0.5s, how far is the ship from the cliff?

I don't understand what to do.
 
  • #7
It's conceptually the same as the first one. Treat it the same way. What is an echo?
 
  • #8
phyzguy said:
It's conceptually the same as the first one. Treat it the same way. What is an echo?

An echo is a repeated sound caused by the reflection of sound waves from a surface. I still don't get what to do :(

Should I multiply 220 by the 0.5? :/
 
Last edited:
  • #9
An echo means the sound went from the ship to the cliff, was reflected from the cliff and traveled back to the ship, where it was heard. Does this help?
 
  • #10
phyzguy said:
An echo means the sound went from the ship to the cliff, was reflected from the cliff and traveled back to the ship, where it was heard. Does this help?

I understand that but I just don't know how to work it out.. would the distance be the same?
 
  • #11
How far does the sound have to go to go from the ship to the cliff and back?
 
  • #12
phyzguy said:
How far does the sound have to go to go from the ship to the cliff and back?

220metres...
 
  • #13
If the ship is 220 meters from the cliff, how far does the sound have to go to go from the ship to the cliff and back?
 
  • #14
440 metres?
 
  • #15
OK, good. Now how fast is the sound traveling (remember it is in air)? And given this, how long will it take?
 
  • #16
1] this was posted elsewhere and,
2] taken as-is, it's a trick question. The answer is provided in the question.
 

1. What is the equation for finding depth given time only?

The equation for finding depth given time only is d = (1/2)gt^2, where d is the depth, g is the acceleration due to gravity, and t is the time.

2. How does this equation work?

This equation works by using the relationship between distance, time, and acceleration due to gravity. It assumes that there is no initial velocity and only the force of gravity acting on the object.

3. Can this equation be used for any type of object?

Yes, this equation can be used for any object that is falling freely due to the force of gravity. However, it does not take into account air resistance or other external forces.

4. What units should be used for the values in this equation?

The units for distance should be in meters (m), time should be in seconds (s), and acceleration due to gravity is typically measured in meters per second squared (m/s^2).

5. Are there any limitations to using this equation?

Yes, there are some limitations to using this equation. It assumes a constant acceleration due to gravity, neglects air resistance, and does not take into account initial velocity or other external forces. It is best used for objects falling near the Earth's surface.

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