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eelshaikh
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Homework Statement
A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of 0.50 s, the top-to-bottom height of the window is 2.00 m. How high above the window top does the flowerpot go?
Homework Equations
Constant acceleration equations
The Attempt at a Solution
The answer in the back of the textbook is listed as 2.34 m above the window top. However, using the following technique:
v (velocity at the top)
v= v0 + at
0= v0 + at
0 - at = v0
0 - (-9.8 m/s^2)(.5) = v0
4.9 m/s = v0
v^2 = v0^2 + 2a(x-x0)
0 = v0^2 + 2g(y-y0)
-v0^2 = 2gy
-v0^2/2g = y
= 1.225 m
which apparently is inconsistent with the text's answer. Help!