\(\displaystyle{y}-{y}_{{0}}={v}_{{0}}{t}+{\frac{{{1}}}{{{2}}}}{>}^{{2}}\)

\(\displaystyle{v}_{{0}}={\frac{{{y}-{y}_{{0}}-{\frac{{{1}}}{{{2}}}}{>}^{{2}}}}{{{t}}}}\)

\(\displaystyle{v}_{{0}}={\frac{{{\left({1.9}\right)}-{\left({4.9}\times{0.38}^{{2}}\right)}}}{{{0.38}}}}={3.138}\ \frac{{m}}{{s}}\)

The flowerpot falls from rest and we have the velocity it gains until it reaches the top of the window below, from this infirmation we can get the distance between the sill and the top of the window below.

\(\displaystyle{v}^{{2}}={{v}_{{0}}^{{2}}}+{2}{g{{\left({y}-{y}_{{0}}\right)}}}\)

\(\displaystyle{y}-{y}_{{0}}={\frac{{{v}^{{2}}-{v}_{{0}}}}{{{2}{g}}}}={\frac{{{3.138}^{{2}}-{0}}}{{{2}\times{9.8}}}}={0.5}\) m

Result: The top of the windows is 0.5 m below the sill of the upper window