Question 17538
Let me begin by giving you an A+ for patience and persistence in waiting for a solution to this problem.   Let me take a shot at it for you.


Let x = the depth of the well and 
let t = time that it takes the rock to reach the bottom of the well.
6-t = time of the sound to return to the top of the well.


(Since the total time to hear the sound is 6 seconds, then the time it takes the sound of the rock hitting the water to return to the top of the well is 6-t.)


s = 16t^2 down, due to gravitational constant 32 ft/sec^2  


D=RT up, where R= 1100 ft/sec (speed of sound) and T= (6-t) seconds up.


s down = D up
{{{16t^2 = 1100*(6-t) }}}

{{{16t^2 = 6600-1100t) }}}


Quadratic equation, so set equal to zero by adding 1100t and subtracting 6600 to each side:
{{{16t^2 +1100t- 6600= 0 }}}


Divide both sides by 4 to make the numbers smaller:
{{{4t^2 + 275t - 1650=0 }}}


Use the quadratic formula to solve for t where of course, t>0.  You may disregard the negative answer to this, since t represents time:

To do this use a pluggable solver, courtesy of algebra.com guru I. Chudov:

*[invoke quadratic "x", 4, 275, -1650]


After finding that the value of t is approximately 5.55169 seconds, there are two ways to calculate the depth of the well.  Let's do it both ways to check:

s= 16t^2 = 16(5.55169)^2 = 493 feet (approximately!)
D= RT = 1100(6-5.55169) =  493 feet (approximately!)


I hope you have a calculator for this!! 

 
R^2 at SCC