Answer
The flowerpot goes up to a height of $~~2.34~m~~$ above the top of the window.
Work Step by Step
Let $v_t$ be the speed at the top of the window.
Let $v_b$ be the speed at the bottom of the window.
Note that the flowerpot is in view for $0.25~s$ on the way up and $0.25~s$ on the way down.
Then:
$v_b = v_t+at$
$v_b = v_t+(9.8)(0.25)$
$v_b = v_t+2.45$
Let $v_{ave}$ be the average speed while passing the window.
We can find $v_t$:
$v_{ave}~t = \Delta y$
$\frac{v_t+v_b}{2} = \frac{\Delta y}{t}$
$\frac{v_t+v_t+2.45}{2} = \frac{\Delta y}{t}$
$v_t+1.225 = \frac{\Delta y}{t}$
$v_t = \frac{\Delta y}{t}-1.225$
$v_t = \frac{2.00}{0.25}-1.225$
$v_t = 6.775~m/s$
Let's consider the flowerpot dropping from maximum height with $v_0=0$
We can find $y$, the maximum height the flowerpot reaches above the top of the window:
$v_t^2 = v_0+2ay$
$v_t^2 = 0+2ay$
$y = \frac{v_t^2}{2a}$
$y = \frac{(6.775~m/s)^2}{(2)(9.8~m/s^2)}$
$y = 2.34~m$
The flowerpot goes up to a height of $~~2.34~m~~$ above the top of the window.
