Determine depth of a well using the speed of sound

[slurik]

Homework Statement

If you drop a stone into a deep well and hear a splash 4.68s after dropping the stone, how far down is the water level? Neglect air resistance and assume that the speed of the sound in air is 3.40x10^2 m/s

Homework Equations

v=d/t
d=v1t+1/2at^2

The Attempt at a Solution

Here is what I had a shot at:

for sound:
v=d/t
d=vt
d=340t

for the stone:
d=v1t+1/2at^2
d=0(4.86s)+1/2(-9.8m/s^2)(4.86-t)^2

for the system:
d=d
so:
340t=1/2(-9.8m/s^2)(4.68-t)^2
340t=-4.9m/s^2 (t^2-9.36t+21.9024)
340t=-4.9t^2+45.864t-107.32176
0=-4.9t^2-294.136t-107.32176

Using that quadratic, I substitute it into the quadratic formula and obtain 2 extraneous solutions:
t=-59.6606...s or t=-0.367sI can see that the amount of time it will take the sound to reach the observers ear is very small in comparison to the amount of time it will take the rock to fall to the bottom. as such, I expect t to be very small since sound is traveling at 340m/s. also I noticed that 1/2at^2 is = approx 107m. The best result I have come up with is 124m, which exceeds the above mentioned result.

The solution sheet says 95.0m

*slams head into desk*

Please help

 

[vela]

You have a sign mistake. The lefthand side is positive when t>0 while the righthand side is negative because you used a=-9.8 m/s^2.

 

[slurik]

Gah! Thanks so much. Sometimes signs get the better of me when I've been working a long time. I have found the solution. Many thanks again! Time for a break perhaps.