Step 1

The given impulse response

\(\displaystyle{h}{\left({t}\right)}={5}{e}^{{-{3}{t}}}{u}{\left({t}\right)}\)

Take Inverse Laplace

\(\displaystyle{H}{\left({s}\right)}={\frac{{{5}}}{{{s}+{3}}}}\)

Step 2

\(\displaystyle{y}{\left({t}\right)}={h}{\left({t}\right)}\cdot{u}{\left({t}\right)}\)

Convert in s domain

\(\displaystyle{Y}{\left({s}\right)}={H}{\left({s}\right)}{U}{\left({s}\right)}\)

\(\displaystyle{Y}{\left({s}\right)}={\frac{{{5}}}{{{s}{\left({s}+{3}\right)}}}}\)

Take partial fractions

\(\displaystyle{Y}{\left({s}\right)}={\frac{{\frac{{5}}{{3}}}}{{{s}}}}+{\frac{{-\frac{{5}}{{3}}}}{{{s}+{3}}}}={\frac{{{5}}}{{{3}}}}{\left({\frac{{{1}}}{{{s}}}}-{\frac{{{1}}}{{{s}+{3}}}}\right)}\)

\(\displaystyle{y}{\left({t}\right)}={\frac{{{5}}}{{{3}}}}{\left({1}-{e}^{{-{3}{t}}}\right)}\)

\(\displaystyle{y}{\left({0.6}\right)}={\frac{{{5}}}{{{3}}}}{\left({1}-{e}^{{-{3}\times{0.6}}}={1.4}\right.}\)

The given impulse response

\(\displaystyle{h}{\left({t}\right)}={5}{e}^{{-{3}{t}}}{u}{\left({t}\right)}\)

Take Inverse Laplace

\(\displaystyle{H}{\left({s}\right)}={\frac{{{5}}}{{{s}+{3}}}}\)

Step 2

\(\displaystyle{y}{\left({t}\right)}={h}{\left({t}\right)}\cdot{u}{\left({t}\right)}\)

Convert in s domain

\(\displaystyle{Y}{\left({s}\right)}={H}{\left({s}\right)}{U}{\left({s}\right)}\)

\(\displaystyle{Y}{\left({s}\right)}={\frac{{{5}}}{{{s}{\left({s}+{3}\right)}}}}\)

Take partial fractions

\(\displaystyle{Y}{\left({s}\right)}={\frac{{\frac{{5}}{{3}}}}{{{s}}}}+{\frac{{-\frac{{5}}{{3}}}}{{{s}+{3}}}}={\frac{{{5}}}{{{3}}}}{\left({\frac{{{1}}}{{{s}}}}-{\frac{{{1}}}{{{s}+{3}}}}\right)}\)

\(\displaystyle{y}{\left({t}\right)}={\frac{{{5}}}{{{3}}}}{\left({1}-{e}^{{-{3}{t}}}\right)}\)

\(\displaystyle{y}{\left({0.6}\right)}={\frac{{{5}}}{{{3}}}}{\left({1}-{e}^{{-{3}\times{0.6}}}={1.4}\right.}\)