1. The problem statement, all variables and given/known data A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of 0.50 s, the top-to-bottom height of the window is 2.00 m. How high above the window top does the flowerpot go? 2. Relevant equations Constant acceleration equations 3. The attempt at a solution The answer in the back of the textbook is listed as 2.34 m above the window top. However, using the following technique: v (velocity at the top) v= v0 + at 0= v0 + at 0 - at = v0 0 - (-9.8 m/s^2)(.5) = v0 4.9 m/s = v0 v^2 = v0^2 + 2a(x-x0) 0 = v0^2 + 2g(y-y0) -v0^2 = 2gy -v0^2/2g = y = 1.225 m which apparently is inconsistent with the text's answer. Help!!!