Time of flight \(T=\frac{2 u \sin \theta}{g}=\frac{2(20 \sqrt{2}) \frac{1}{\sqrt{2}}}{10}=4\) second
\(\Rightarrow\) After \(2\) second particle will be at maximum height of the projectile \(L=m v r_{\perp}\)
\(r_{\perp}=H_{\max }=\frac{u^2 \sin ^2 \theta}{2 g}=20 \,m\)
So \(L=(1)(20)(20)=400(-\hat{k})\)