Depth of a well (using a stone and sound)

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SUMMARY

The discussion focuses on calculating the depth of a well using the time it takes for a stone to hit the bottom and the sound of the splash to travel back up. The total time recorded is 2.00 seconds, with the speed of sound at 10.0 degrees Celsius calculated as 337.0 m/s. The equation used combines the time for the stone to fall and the time for the sound to travel, leading to the equation 2.0 s = x/337 + sqrt(2x/9.8). The correct depth of the well is determined to be 18.5 meters, and the solution can be simplified to a quadratic equation for easier resolution.

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Homework Statement


A stone is dropped from rest into a well. The sound of the splash is heard exactl 2.00 s later. Find the depth of the well if the air temperatrue is 10.0 degrees Celsius.


Homework Equations


Possibly the Kinematics Equation
V= Vo sqrt(1+Temp/273)


The Attempt at a Solution


First i found the speed of the sound and came up with
v = 337.0 m/s

i used the kinematics equation to find the time it took for the stone to hit the bottom of the well and it is

Tstone = sqrt(2x/g)

so i the total time equals the time it takes for the stone to hit the ground and the sound to come up so

2.0 s = x/337 + sqrt( 2x/9.8 ) where x equals the depth of the well

I don't think it is right, and I can't seem to solve for x. Can anyone help? The answer should be 18.5 m
 
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Looks correct to me! Kind of a messy equation, though.
I used to give that question to my grade 11 physics class so they would have to learn the method of trial and error (guess and test). You can zero in on the right distance in 4 or 5 trials. Too bad you already know the answer.

If you take the x/337 term to the other side so the root is isolated on one side, then square both sides, you will have a quadratic equation. You will no doubt know how to solve a quadratic without resorting to trial and error.
 
Looks OK to me.
Your equation is a quadratic in sqrt(x). Let x = y^2
 

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