Angular Mechanics + Kinematics

Let the co-ordinates of $P$ be $(0,0)$ and those at a point $Q$ where the body is at $t$ $=$ $(\frac{v_o}{g}$ ) be taken for granted as $(x,y)$ $.$

Horizontal component of velocity is $v_o$ $cos45^o$ which is $(\frac{v_o}{\sqrt{2}}$ )

Similarly vertical component of velocity is $v_o$ $sin45^o$ which is also $(\frac{v_o}{\sqrt{2}}$ )

$x$ which is the horizontal distance covered in time $t$ $=$ $(\frac{v_o}{g}$ ) is the horizontal component of velocity times time taken. This is $\frac{v_o^2}{g\sqrt{2}}$

And $y$ the vertical distance covered is $(\frac{v_o^2}{2g}$ ) $\sqrt{2}-1$

Horizontal velocity $v_x$ at $Q$ is $(\frac{v_o}{\sqrt{2}}$ )

Vertical velocity $v_y$ at $Q$ is $\frac{1-\sqrt{2}}{\sqrt{2}}$ $v_o$

Angular momentum at $Q$ about $P$ is given by $m$ $(xv_y - yv_x)$

Substituting and simplifying we get $L$ $=$ $-$ $(\frac{mv_o^3}{2g\sqrt{2}}$ )

Finally substituting we get $L$ $=$ $-0.0357$ which is $\approx$ $-0.04$

Taking only magnitude we get $4$ $\times$ $10^-2$ .