Physics
posted by Waqas .
A drowsy cat spots a flower pot that sails first up and then down past an open window. The pot is in view for a total of 0.50s. And the top to bottom height of the window is 2.00m. How high above the window top does the flower pot go?

You can simplify the problem a bit by realizing that the pot will take the same length of time going up as going down past the window's 2 meter height. Thus it takes 0.25 s to go from top to bottom. Let H be the verticaldistance from maximum height to the top of the window. Let t1 be the time it takes to go from maximum height H to the top of the window, and t2 is the time it takes to go from maximum height H to the bottom of the window,
t2  t1 = 0.25 s
(1/2)g t1^2 = H
(1/2)g t2^2 = H + 2
(1/2)g (t1 + 0.25)^2 = H + 2
(1/2)g (t1 + 0.25)^2  (1/2)t1^2 = 2
You now have only one unknown. Solve for t1 and with that you can solve for H. 
1.571556 meters