SOLUTION: You drop a rock into a deep well. You can't see the rock's impact at the bottom, but you hear it after 6 seconds. The depth of the well is ____feet. Ignore air resistance. The ti

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Question 17538: You drop a rock into a deep well. You can't see the rock's impact at the bottom, but you hear it after 6 seconds. The depth of the well is ____feet.
Ignore air resistance. The time that passes after you drop the rock has two components: the time it takes the rock to reach the bottom of the well, and the time that it takes the sound of the impact to travel back to you. Assume the speed of sound is 1100 feet per second.

Answer by rapaljer(4667) About Me  (Show Source):
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
16t%5E2+=+1100%2A%286-t%29+
16t%5E2+=+6600-1100t%29+

Quadratic equation, so set equal to zero by adding 1100t and subtracting 6600 to each side:
16t%5E2+%2B1100t-+6600=+0+

Divide both sides by 4 to make the numbers smaller:
4t%5E2+%2B+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:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 4x%5E2%2B275x%2B-1650+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28275%29%5E2-4%2A4%2A-1650=102025.

Discriminant d=102025 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-275%2B-sqrt%28+102025+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28275%29%2Bsqrt%28+102025+%29%29%2F2%5C4+=+5.5516906342111
x%5B2%5D+=+%28-%28275%29-sqrt%28+102025+%29%29%2F2%5C4+=+-74.3016906342111

Quadratic expression 4x%5E2%2B275x%2B-1650 can be factored:
4x%5E2%2B275x%2B-1650+=+4%28x-5.5516906342111%29%2A%28x--74.3016906342111%29
Again, the answer is: 5.5516906342111, -74.3016906342111. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+4%2Ax%5E2%2B275%2Ax%2B-1650+%29



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