A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot was in view for a total of 0.61 seconds, and the top-to-bottom height of the window is 2.45 meters. How high above the window top did the flowerpot go? I first tried to calculate the initial velocity by assuming the initial position to be zero and the 2.45 for the max. Using formula y=-1/2gt^2 + v(t) +Yinitial 2.45 = -1/2g(0.61)^2 + V(0.61) o intial velocity=7.00 m/s velocity should equal zero, thats when the flowerpot reached its highest point. Then solving for time when the pots velocity is zero, will be give enough info to calculate the height above the window the pot traveled. Using equation v=-gt+vintial 0= -gt + 7.00 t= 0.714s Plugging back into the first used equation y= -4.9(0.714)^2 +7.00(0.714) y= 2.50m subtracting from 2.45m = 0.049m So the flowerpot went 0.049m above the window??? I enter my homework answers on this site called webassign. My first answer was incorrect and another wrong answer will start deducting points.... I am really confused because my textbook had the same Homework problem, however the values were different. height was 2.00m, time= 0.50s. I did the same procedure as stated above on the problem in my text, but i could Not get close to the answer of 2.34m from the solution in the back of the book.