Depth of a well (using a stone and sound)

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A stone is dropped into a well, and the splash sound is heard after 2.00 seconds, with the air temperature at 10.0 degrees Celsius. The speed of sound is calculated to be 337.0 m/s. The total time includes the fall time of the stone and the time for the sound to travel back up, leading to the equation 2.0 s = x/337 + sqrt(2x/9.8). The discussion suggests isolating the square root term to form a quadratic equation for easier solving. The expected depth of the well is 18.5 meters, and the approach involves trial and error or solving the quadratic directly.
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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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