To solve this problem we will use the linear motion kinematic equations for which acceleration (in this case gravity) is described as the change in velocity in an instant of time. Subsequently through the energy conservation equations we will find the net height.
Let's start considering that the time is 1 second, therefore,
[tex]t = 1s[/tex]
Now the acceleration would be given as
[tex]g = \frac{\Delta v}{t}[/tex]
[tex]g = \frac{v_f-v_i}{t}[/tex]
[tex]v_f-v_i = gt[/tex]
Since the final speed is zero, and the time elapsed to reach the height is 0.5 seconds (half of the entire route) we will have that the initial speed can be expressed in terms of gravity as
[tex]0-v_i = g(\frac{1}{2})[/tex]
[tex]v_i = \frac{1}{2} g[/tex]
Now for energy conservation, the potential energy must be equal to the kinetic energy, therefore,
PE = KE
[tex]mgh = \frac{1}{2} mv_i^2[/tex]
[tex]h = \frac{v_i^2}{2g}[/tex]
Replacing the previous value found,
[tex]h = \frac{1}{4} \frac{g^2}{2g}[/tex]
[tex]h = \frac{1}{4} \frac{9.8^2}{2(9.8)}[/tex]
[tex]h = 1.225m[/tex]
Therefore the hegiht above the windows that the pot rises is 1.225m