Solve Drowsy Cat Flower Pot Height Question

[rootX]

Homework Statement

A drowsy cat spots a flower pot that sails first up the window, and then down past an open window. The pot is in view for a total of 0.5s, and top to bottom height of the window is 2 m. How high above the window top does the flower pot go?

Homework Equations

Using kinematics equations:
h = v.t + 0.5*a*t^2
v[f]-v = at
or, 2h/t = v[f]+v

The Attempt at a Solution

I assumed that it is going down, while it was viewed.
so, here's my diagram:

0
^|
^|
^s
^|
^|
v
^|
^|
^2 m, 0.5 s
^|
v[f]

(motion is opposite to the direction indicated by the carrot symbols)

so using those equations I get
v = 1.5475 m/s

and as 2as = v^2

so s = 0.122 m

but the answer at the back of the book is 2.34 m ><

 

[rootX]

oops, I got it!
I was not paying enough attention to the wording of this question.
"for a total of 0.5s"

I should have read this thread before posting this one:
https://www.physicsforums.com/showthread.php?t=56362

 

[m2003]

I would suggest that there may be some missing information in this problem. Without knowing the initial and final velocities of the flower pot, it is impossible to accurately solve for its height above the window top. Additionally, the equation used in the attempt at a solution is for a constant acceleration, which may not necessarily apply to this situation. It would be helpful to have more information, such as the weight of the flower pot, any external forces acting on it, and the angle at which it was thrown. Without these details, it is difficult to provide a definitive answer.