Posted by **Kristen** on Sunday, September 9, 2007 at 6:08pm.

A drowsy cat spots a flowerpot that saild first up and then down past an open window. The pot was in view for a total of .43 s, and then top-to-bottom height of the window is 2m. How high above the window top did the flowerpot go?

- Physics -
**drwls**, Sunday, September 9, 2007 at 9:59pm
Let V2 be the velocity as the pot goes by the bottom of the windowm and V1 be the velocity as it goes by the bottom. The time-averaged velocity as it passes by is (V1 + V2)/2 = 2/0.43 = 4.65 m/s

The change in velocity as it goes by is (V1-V2)/2 = gt = 9.8 m/s^2*0.43 = 0.42 m/s

You now have two equations that let you solve for both V1 and V2

V1 + V2 = 9.30 m/s

V1 - V2 - 0.42 m/s

2 V2 = 8.88 m/s

V2 = 4.44 m/s

The height H above the window that the pot travels can be obtained by setting the kinetic energy at V2 equal to the gain of potential energy at the highest point:

M g H = (1/2) M V2^2

H = V2^2/2g = 1.01 meter

- Physics -
**Kristen**, Sunday, September 9, 2007 at 10:04pm
thank you SOOOO much, i spent a long time trying to figure this out

- Physics -
**drwls**, Sunday, September 9, 2007 at 11:35pm
I made a mistake typing one sentence and equation. A "1/2" factor should not have been in the velocity charge equation. It should have read

<<The change in velocity as it goes by is (V1-V2)= gt = 9.8 m/s^2*0.43 = 0.42 m/s>>

It does not affect the answer because I ignored the "/2" when doing the numbers

- Physics -
**YOUREWRONG**, Wednesday, January 18, 2012 at 5:03pm
YOUREWRONG

- Physics -
**Anonymous**, Sunday, September 2, 2012 at 3:27pm
I don't understand how you get .42 from 9.8*.43

Every time I do this I get 4.2 m/s and I would like to know what I am doing wrong or if you are making a mistake.

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