How long does it take for the sound to travel a known distance in a well?

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The discussion revolves around calculating the depth of a well based on the time it takes for a stone to hit the bottom and for the sound of the impact to travel back. The total time of 16 seconds includes both the falling time of the stone (t1) and the time for sound to return (t2). The speed of sound is given as 343.8 m/s, while gravity accelerates the stone at 9.8 m/s². Participants emphasize the need to express t1 and t2 in terms of the well's depth and clarify the problem statement for accurate calculations. The conversation highlights the importance of understanding the relationship between the distances traveled by the stone and the sound.
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Homework Statement



How deep is the well? Drop a stone and I took 16s to hit the ground level.

The air temperature is 20 degrees.

Acceleration is 9.8 m/s squared.

M=1kg

V1=0 m/s

Speed of sound= 343.8m/s squared
Find how deep the well is.

Homework Equations



D=t/v

I don't really know at all

Speed of sound.



The Attempt at a Solution



No ideas.
 
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Confusedkid34 said:

Homework Statement



How deep is the well? Drop a stone and I took 16s to hit the ground level.

The air temperature is 20 degrees.

Acceleration is 9.8 m/s squared.

M=1kg

V1=0 m/s

Speed of sound= 343.8m/s squared
Find how deep the well is.

Homework Equations



D=t/v

I don't really know at all

Speed of sound.

The Attempt at a Solution



No ideas.
Hello Confusedkid34. Welcome to PF!

Your obviously missing some words in your problem statement. What is it that takes 16 seconds to hit the ground?

To make all this clearer, please state the problem word for word as it was given to you.
 
Confusedkid34 said:

Homework Statement



How deep is the well? Drop a stone and I took 16s to hit the ground level.

The air temperature is 20 degrees.

Acceleration is 9.8 m/s squared.

M=1kg

V1=0 m/s

Speed of sound= 343.8m/s squared
Find how deep the well is.

Homework Equations



D=t/v

I don't really know at all

Speed of sound.



The Attempt at a Solution



No ideas.

Unless there's more to the problem statement than you've posted, only some of the given information is relevant.

Does the stone have a constant speed as it falls, or does it accelerate? What formula applies?
 
The speed is constant, it takes 16s to hear the Rock hit the bottom.

So 16s for the rock to hit the bottom and for the noise to travel back.

It starts at rest. Both the sound and rock travel the same distance.

Speed of sound is 343.8 m/s

So time equals 16s. t=t1+t2

T1= time for the stone to hit the ground.

T2= time for sound to travel back.

So we need to solve for d. By finding out what t1 and t2 are.
 
Last edited:
Confusedkid34 said:
The speed of sound is constant, it takes 16s to hear the Rock hit the bottom.

So 16s for the rock to hit the bottom and for the noise to travel back.

It starts at rest. Both the sound and rock travel the same distance.

Speed of sound is 343.8 m/s

So time equals 16s. t=t1+t2

T1= time for the stone to hit the ground.

T2= time for sound to travel back.

So we need to solve for d. By finding out what t1 and t2 are.
How far does sound travel in a time of t2 ?

How far does the rock fall in a time of t1 ?

If t1 + t2 = 16, then express t2 in terms of t1 .

Put all of that together, solving for t1.
 
The distance for t1 and t2 are the same.

But I don't understand how to do that?

Somewhere do I have to apply this in a quadratic ?
 
Confusedkid34 said:
The speed is constant, it takes 16s to hear the Rock hit the bottom.

So 16s for the rock to hit the bottom and for the noise to travel back.
What happened to gravity? Your problem statement didn't mention anything about hearing the rock hit the bottom, only that it took 16 second to hit the bottom, no sound involved.

So really your problem statement should go something like:

An observer drops a stone from rest into a deep well. The stone falls under the influence of gravity. 16 seconds after the stone begins its fall the observer hears the sound of the rock hitting the bottom. How deep is the well?
It starts at rest. Both the sound and rock travel the same distance.

Speed of sound is 343.8 m/s

So time equals 16s. t=t1+t2

T1= time for the stone to hit the ground.

T2= time for sound to travel back.

So we need to solve for d. By finding out what t1 and t2 are.

Sounds like a plan. Can you write the expressions for each time?
 
Confusedkid34 said:
The distance for t1 and t2 are the same.

But I don't understand how to do that?

Somewhere do I have to apply this in a quadratic ?
Please use the QUOTE button to indicate which post you are relying to.

It will help a lot.

I'll do this post, then let gneill finish up. (If he's willing.)

You didn't answer those questions I asked in post #5 .

Answering them will help significantly in solving this problem.
 
Your right. I didn't include all the information.

So we need an equation to solve for d. As well we need to figure out t1 and t2.

I have no idea how to do this and I could use some guidance.
 
  • #10
Confusedkid34 said:
Your right. I didn't include all the information.

So we need an equation to solve for d. As well we need to figure out t1 and t2.

I have no idea how to do this and I could use some guidance.

Suppose you happened to know the distance d. How would you write an expression for the amount of time, call it t2, it takes for the sound to travel that far?
 
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