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?

Answer 1

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

Answer 2

thank you SOOOO much, i spent a long time trying to figure this out

Answer 3

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

Answer 4

YOUREWRONG

Answer 5

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.