Posted by **Waqas** on Saturday, September 13, 2008 at 11:36pm.

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**, Sunday, September 14, 2008 at 12:39am
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**, Friday, January 30, 2015 at 11:28pm
1.571556 meters

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