\(\text { Momentum }=\frac{m v}{\sqrt{2}}\)
\(L=\) Angular momentum \(=\) Momentum \(\times\) Perpendicular distance \(L=\frac{m v}{\sqrt{2}(h)}\)
Here \(h=\frac{v^{2} \sin ^{2} 45^{0}}{2 g}=\frac{v^{2}}{4 g}\)
\(\therefore L=\frac{m v}{\sqrt{2}} \frac{v^{2}}{4 g}=\frac{m v^{3}}{4 \sqrt{2} g}\)