Calculating Well Depth Using the Speed of Sound

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Homework Help Overview

The discussion revolves around calculating the depth of a well based on the time it takes for a stone to hit the water and for the sound of the splash to reach the observer. The problem involves concepts of acceleration due to gravity and the speed of sound.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the need to account for both the time it takes for the stone to fall and the time for the sound to travel back. Some suggest focusing on gravitational acceleration, while others emphasize the importance of sound propagation.

Discussion Status

There are differing opinions on whether the speed of sound should be considered in the calculations. Some participants provide guidance on how to approach the problem, while others express confusion about the relevance of sound in the context of the assignment.

Contextual Notes

Participants note that this is a grade K-12 homework problem, which may influence the complexity of the expected solution. There is also mention of potential misunderstandings regarding the problem's focus.

ramollari
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A stone is thrown into a well and the sound is heard after 3s. How deep is the well?
 
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Write down the times the stone takes to hit the water and the sound takes to reach your ear.You'll find your answer in no time...

Daniel.
 
You may also want to factor in the acceleration of the stone due to gravity.
 
Dex, you silly!

This is an assignment about acceleration, not about the speed of sound!
 
DaveC426913 said:
Dex, you silly!

This is an assignment about acceleration, not about the speed of sound!

:mad:

Coul you explain both of your sentences? :mad:

Daniel.
 
Use 343 m/s for the speed of sound, and remember the time given has the time when the rock falls and then the sound wave gets to you. Also the sound will be moving with constant speed.
 
dextercioby said:
:mad:

Coul you explain both of your sentences? :mad:

Daniel.

I believe you missed the forest for the trees.

Your response seems to focus on the amount of time it takes the sound to reach your ear from the bottom of the well.

I believe the question is not about the speed of sound, it is about acceleration under gravity. The only relevant time delay concerns how fast the rock falls due to gravity. One should assume the delay due to propagation of sound is not a factor.

Maybe. I could be wrong.

It's grade K-12 homework.
 
DaveC426913 said:
I believe you missed the forest for the trees.

Your response seems to focus on the amount of time it takes the sound to reach your ear from the bottom of the well.

I believe the question is not about the speed of sound, it is about acceleration under gravity. The only relevant time delay concerns how fast the rock falls due to gravity. One should assume the delay due to propagation of sound is not a factor.

Maybe. I could be wrong.

It's grade K-12 homework.


Dave,i believe you need a pair of glasses... :-p My answer focuses on solving the problem:
dextercioby said:
Write down the times the stone takes to hit the water and the sound takes to reach your ear.You'll find your answer in no time...
,not on the speed of sound.It never mentions the speed of sound...:rolleyes:

Since the solver needs to find the depth,a priori unknown,he must include the time of sound propagation,because he has no idea whether it can,or not be neglected.

Daniel.
 
Doing it the "proper" way involves solving a quadratic. Assuming the speed of sound to be 340 m/s, there is an 8.5 % difference from the "easy" answer found by substitution.

I'm with Dexter on this one.
 
  • #10
You don't think the grade K-12 homework problem is more simply stated as:

"How far will an object fall in 3 seconds"

?
 
  • #11
DaveC426913 said:
You don't think the grade K-12 homework problem is more simply stated as:

"How far will an object fall in 3 seconds"

?

Don't u think that in the first 18 years of life one should know how to solve a quadratic equation?? :wink:

Daniel.
 
  • #12
edit: nvm nvm nvm
 
  • #13
d=.5at^2
d=.5(9.8)(3s)^2
d=44.1m


time*acceleration = velocity
time*velocity = distance
 
  • #14
ramollari did it say anything about the speed of sound in this question?
 

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