A particle of mass 100g is projected at time t=0 with a speed 20 ${\mathrm{ms}}^{-1}$ at an angle 45⁰ to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time t=2 s is found to be $\sqrt{\mathrm{K}} {\mathrm{kgm}}^2/\mathrm{s}$ . The value of $\mathrm{K}$ is ___________.
(Take g=10m/s²)

$m=100\mathrm{g}=0.1\mathrm{kg}$
After time ‘t’, position vector of particle is
$\begin{array}{l}\overrightarrow{\mathrm{r}}=\mathrm{x}\widehat{\mathrm{i}}+\mathrm{y}\widehat{\mathrm{j}}\\ \overrightarrow{\mathrm{r}}=(\mathrm{ucos}\mathrm{\theta })\mathrm{t}\widehat{\mathrm{i}}+\left(\mathrm{usin}\mathrm{\theta t}-\frac{1}{2}{\mathrm{gt}}^2\right)\widehat{\mathrm{j}}\end{array}$
Velocity of particle is
$\begin{array}{l}\overrightarrow{\mathrm{V}}={\mathrm{V}}_{\mathrm{x}}\widehat{\mathrm{i}}+{\mathrm{V}}_{\mathrm{y}}\widehat{\mathrm{j}}\\ \overrightarrow{\mathrm{V}}=\mathrm{ucos}\mathrm{\theta }\widehat{\mathrm{i}}+(\mathrm{usin}\mathrm{\theta }-\mathrm{gt})\widehat{\mathrm{j}}\end{array}$
Angular momentum of particle
$\begin{array}{l}\overrightarrow{\mathrm{L}}=\mathrm{m}(\overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{v}})\\ \overrightarrow{\mathrm{L}}=\mathrm{m}\left[{\mathrm{u}}^2\sin \mathrm{\theta cos}\mathrm{\theta t}-{\mathrm{u}}^2\sin \mathrm{\theta cos}\mathrm{\theta t}+\frac{1}{2}{\mathrm{gt}}^2\mathrm{ucos}\mathrm{\theta }\right]\widehat{\mathrm{k}}\\ \left|\overrightarrow{\mathrm{L}}\right|=\frac{1}{2}{\mathrm{mgt}}^2\mathrm{ucos}\mathrm{\theta }\\ \left|\overrightarrow{\mathrm{L}}\right|=\frac{1}{2}\times 0.1\times 10\times 4\times 20\times \frac{1}{\sqrt{2}}=\frac{40}{\sqrt{2}}=20\sqrt{2}\\ \left|\overrightarrow{\mathrm{L}}\right|=\sqrt{800} {\mathrm{kgm}}^2/\mathrm{s}\end{array}$