For an "exact" solution, we can also use the following variant of the argument of Henning Makholm.
Let the height above the window be $h$, let the total time to reach the top of the window be $t_1$, and total time to reach the bottom of the window be $t_2$. Let acceleration due to gravity be constant at $a=9.8$.
Then
$$h=\frac{1}{2}at_1^2 \qquad\text{and}\qquad h+1.35=\frac{1}{2}at_2^2.$$
Subtract. We get
$$\frac{1}{2}a(t_2^2-t_1^2)=1.35.$$
But $t_2^2-t_1^2=(t_2-t_1)(t_2+t_1)$. Since $t_2-t_1=0.210$ we obtain
$$t_2+t_1=\frac{(2)(1.35)}{(0.210)(9.8)}.$$
Note that $t_1=(1/2)((t_2+t_1)-(t_2-t_1))$. Use our expression for $t_2+t_1$, together with $t_2-t_1=0.210$, to find $t_1$. Now we know $t_1$, so we know $h$.
Another way: It is more pleasant to avoid numbers until the end. Let $w$ be the height of the window, and let $s$ be the amount of time it took for the object to pass the window. Let $u$ be the velocity at the top of the window, and
$v$ the velocity at the bottom. Then the average velocity at which the window was traversed is $(v+u)/2$. But it is also $w/s$, and therefore
$$\frac{v+u}{2}=\frac{w}{s}.$$
The change in velocity is $v-u$. It is also $as$. Thus
$$\frac{v-u}{2}=\frac{as}{2}.$$
From the above two equations it follows that
$$u=\frac{w}{s}-\frac{as}{2}.$$
The average velocity from the time of dropping until arriving at the window is $u/2$. The time it took is $u/a$, so the distance travelled is $u^2/2a$. It follows that the height above the window from which the object was dropped is
$$\frac{\left(\frac{w}{s}-\frac{as}{2}\right)^2}{2a}.$$
Assume upwards is positive. We have two equations:
$$\begin{align}
-y =& \frac{1}{2}(-9.81)t^2\\
-(y+1.90) =& \frac{1}{2}(-9.81)(t+.420)^2
\end{align}$$
where $t$ is the time until the flowerpot reaches the top of the lower window. You should be able to find out this $t$ and then get $y$.
Best Answer
Don't know if the question is on-topic for https://math.stackexchange.com/ but the approach will be as follows:
Let initial velocity be $ \vec{u} $ towards the positive y-direction (up). Acceleration due to gravity is $ - \vec{g} $ towards negative y-direction (down).
Use Newton's equations of linear motion:
$$ \begin{align} \vec{v} &= \vec{u} + t * \vec{a} \\ \vec{s} &= t * \vec{u} + \frac{1}{2} t^2 * \vec{a} \\ v^2 &= u^2 + 2 * a * s \end{align} $$
You'll only need the second equation for two different values of $ t $.