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 toptobottom 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 timeaveraged 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 (V1V2)/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 (V1V2)= 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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