1. The problem statement, all variables and given/known data A stone is dropped in a well. The splash is heard 3s later. What is the depth of the well? 2. Relevant equations constant acceleration equations take speed of sound to be 343 m/s. 3. The attempt at a solution For the stone: u = 0 a = g s = ut + 0.5at^2 subbing it all in you get t_stone = +sqrt(2s/g) For the sound wave: u = v = 343 m/s v = s/t subbing it all in you get t_wave = s/343 (same distance travelled for stone and wave = s) so you have t_stone + t_wave = 3 therefore +sqrt(2s/g) + s/343 = 3 squaring both sides and moving over the constant (s^2/343^2) + 2s/g - 9 = 0 solving this quadratic I get s = 44.02m or -24054m The answer is 40.7m. Where did I go wrong?