Ok I am at this problem for hours and I am still nowhere, I am just wondering if any of you can help. Here is the question A rock is dropped from rest into a well. If the sound of the splash is heard 2.42s later, how far below the top of the well is the surface of the water? (take the speed of sound to be 336.0 m/s and take the freefall acceleration of the rock to be 9.8 m/s^2. So I figure ok this must mean 2.42s = Time spent falling + Amount of time it took for the sound to reach the top. So I try to use one of the kinematic equations for motions with constant acceleration Sf = Si + Vi*(Change of T) + 1/2 * A *(Change of T) ^ 2 Since Initial Position Si = 0 and Vi = 0 The formula will look like Sf = 1/2 *A*(Change of T) ^ 2 Since I need to get rid of one of the unknowns I do this Sf = 1/2 * 9.8 * (2.42s - SF/336m/s) ^ 2 Sf = 4.9 * (2.42s- SF/336m/s) ^ 2 Sf = 4.9 * (813.12m - SF)/336m/s)^ 2 And then yeah..I went further then that but let's just say it didn't give me the right answer. I tried several other methods but nothing gave me the right answer. Any Ideas on how to resolve this matter?