Sound and kinematic problem solving

AI Thread Summary
To solve the problem of how long it takes for a man to hear the sound of a stone hitting the bottom of a 120m can, two time periods must be considered: the time for the stone to fall and the time for the sound to travel back up. The initial calculation for the stone's fall time was correct at approximately 7.2 seconds, but it did not account for the sound's travel time. The speed of sound, which is temperature-dependent, needs to be calculated using the given air temperature of 20 degrees Celsius. The final answer requires adding both time periods together to determine the total time until the sound is heard. Understanding these components is crucial for accurately solving the problem.
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Homework Statement


a man dropped a stone into a can that is 120m deep. if the temperature of the air is 20degree celsius. how long would it take for a man to hear the sound when the stone hit the can?

Homework Equations



d=Vi*t + 1/2*a*t^2


The Attempt at a Solution


since i know that the initial velocity is 0.0m/s
i canceled out the part vi*t
and solved for time ----> t= \sqrt{}2(d)/a
which is t= \sqrt{}2(120m)/9.81m/s^2= 7.2s

is this correct.? I am having a doubt on my answer because i didnt use the temperature at all.
 
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I'm not sure how you got 7.2s from that equation - maybe you entered in incorrectly into your calculator?

And the temperature is really only used as a reference in regards to the speed of sound. So, you have the time it takes the stone to hit the bottom of the can, and then the time for the sound to travel back to the man.
 
p21bass said:
I'm not sure how you got 7.2s from that equation - maybe you entered in incorrectly into your calculator?

And the temperature is really only used as a reference in regards to the speed of sound. So, you have the time it takes the stone to hit the bottom of the can, and then the time for the sound to travel back to the man.

wait. sorry i got wrong number from the calculator but the equation is right.right?
and also so are you saying that i had to use the temperature. and find thae speed of the time and add it with the time i got from the kinematic equation or just what i did was the end?
 
p21bass said:
I'm not sure how you got 7.2s from that equation - maybe you entered in incorrectly into your calculator?

And the temperature is really only used as a reference in regards to the speed of sound. So, you have the time it takes the stone to hit the bottom of the can, and then the time for the sound to travel back to the man.

Unless, of course, the man hears the sound that travels up through the walls of the can first! :wink:

That's a *big* can!
 
gneill said:
Unless, of course, the man hears the sound that travels up through the walls of the can first! :wink:

That's a *big* can!

what? wait. I am still confused so you mean the equation i did was the end. or i had to do add the time i got using the temperature?
 
There are two time periods involved. The first (which you've already addressed) is the time it takes for the stone to reach the bottom of the can. The second period is the time it takes for the sound from the impact to travel back up to the man. This second period involves the speed of sound in air which is temperature dependent (but not by too much unless the temperatures get really far from room temp). Simply look up the speed of sound at that temperature.
 
gneill said:
There are two time periods involved. The first (which you've already addressed) is the time it takes for the stone to reach the bottom of the can. The second period is the time it takes for the sound from the impact to travel back up to the man. This second period involves the speed of sound in air which is temperature dependent (but not by too much unless the temperatures get really far from room temp). Simply look up the speed of sound at that temperature.

no!so that means in the end you were supposed to add both of the times and that is your final answer? cause this question was on my test i stopped at the kinematic equation.
 
Yes, that's how that problem is solved.
 
p21bass said:
Yes, that's how that problem is solved.

ah man..at lest i did the half correctly . thank you so much for help! it was a great help!
 
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