Posted by Waqas on .
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?

Physics 
drwls,
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. 
Physics 
Karl Benedict Banag,
1.571556 meters