You can put this solution on YOUR website!
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
Quadratic equation, so set equal to zero by adding 1100t and subtracting 6600 to each side:
Divide both sides by 4 to make the numbers smaller:
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:
|Solved by pluggable solver: SOLVE quadratic equation with variable|
|Quadratic equation (in our case ) has the following solutons:|
For these solutions to exist, the discriminant should not be a negative number.
First, we need to compute the discriminant : .
Discriminant d=102025 is greater than zero. That means that there are two solutions: .
Quadratic expression can be factored:
Again, the answer is: 5.5516906342111, -74.3016906342111.
Here's your graph:
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