- #1

- 4

- 0

## 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!!!