The vertical acceleration

\(\displaystyle{a}_{{y}}={\frac{{-{q}{E}}}{{{m}}}}={\frac{{-{\left({1.6}\cdot{10}^{{-{19}}}\right)}{\left({5}\cdot{10}^{{3}}\right)}}}{{{9.11}\cdot{10}^{{-{31}}}{k}{g}}}}\)

\(\displaystyle=-{8.78}\cdot{10}^{{{44}}}\frac{{m}}{{\sec{}}}\)

The vertical velocity is

\(\displaystyle{{V}_{{y}}^{{2}}}={{V}_{{{o}{y}}}^{{2}}}+{2}{a}_{{y}}{h}\)

\(\displaystyle{V}_{{o}}{\sin{{45}}}={0}+\sqrt{{-{2}{a}_{{y}}{h}}}\)

\(\displaystyle{V}_{{o}}={\frac{{{\left[-{\left(-{8.78}\cdot{10}^{{{14}}}\frac{{m}}{{s}^{{2}}}\right)}{\left({0.010}{m}\right)}\right]}^{{\frac{{1}}{{2}}}}}}{{{\sin{{45}}}}}}\)

\(\displaystyle={4.2}\cdot{10}^{{6}}\frac{{m}}{{s}}\)

\(\displaystyle{a}_{{y}}={\frac{{-{q}{E}}}{{{m}}}}={\frac{{-{\left({1.6}\cdot{10}^{{-{19}}}\right)}{\left({5}\cdot{10}^{{3}}\right)}}}{{{9.11}\cdot{10}^{{-{31}}}{k}{g}}}}\)

\(\displaystyle=-{8.78}\cdot{10}^{{{44}}}\frac{{m}}{{\sec{}}}\)

The vertical velocity is

\(\displaystyle{{V}_{{y}}^{{2}}}={{V}_{{{o}{y}}}^{{2}}}+{2}{a}_{{y}}{h}\)

\(\displaystyle{V}_{{o}}{\sin{{45}}}={0}+\sqrt{{-{2}{a}_{{y}}{h}}}\)

\(\displaystyle{V}_{{o}}={\frac{{{\left[-{\left(-{8.78}\cdot{10}^{{{14}}}\frac{{m}}{{s}^{{2}}}\right)}{\left({0.010}{m}\right)}\right]}^{{\frac{{1}}{{2}}}}}}{{{\sin{{45}}}}}}\)

\(\displaystyle={4.2}\cdot{10}^{{6}}\frac{{m}}{{s}}\)