Physics freefall problem?!!!?

Question

A rock is dropped from rest into a well. (a) If the sound of the splash is heard 2.40s later, how far below the top of the well is the surface of the water? The speed of sound in air (at the ambient temperature) is 336 m/s. (b) IF the travel time for the sound is ignored, what percentage error is introduced when the depth of the well is calculated?

Answer 1

The total time for the rock to drop and the sound to come back is 2.40s. The distance that the rock drops is (.5)(9.8 m/s^2)(time for rock to fall). The distance that the sound comes back is (336 m/s)(time of sound to come back). The two times added together are 2.4.
These two distances are the same so.
(.5)(9.8m/s^2)(time for rock to fall)^2= (336 m/s)(2.4 s - time for rock to fall)
(4.9m/s^2)(time)^2 = 806.4m - (336 m/s)(time for rock to fall)
(4.9m/s^2)(time)^2 +336 m/s(time) -806.4 m = 0
Using the quadratic formula, the allowed values for time is
2.31 s
Distance down to water is (.5)(9.8 m/s^2)(2.31s)^2 = 26.15 m

b) if you were to ignore travel time for the sound, you would simply have (.5)(9.8m/s^2)(2.4s)^2 = 28.22 m. Your percentage error would be (28.22-26.15)/26.15 = 7.9%