We know that angular momentum vecL→L of a particle of mass mm moving with velocity vecv→v with respect to a selected origin is given by the equation
vecL=vecrxxvecp→L=→r×→p
where vecp→p is linear momentum and =mvecv=m→v

The velocity of the particle changes continuously due to acceleration due to gravity gg acting on the vertical component of velocity which =vsin45^@=v/sqrt2=vsin45∘=v√2.
Similarly horizontal component of velocity=vcos45^@=v/sqrt2=vcos45∘=v√2
We know that at maximum height vertical component of velocity is zero.
![enter image source here]()
(Please read initial velocity as vv instead of "u"u in the figure, theta=45^@θ=45∘)
Using the kinematic equation
v^2-u^2=2asv2−u2=2as and inserting given values we get
0^2-(v/sqrt2)^2=2(-g)h02−(v√2)2=2(−g)h
=>h=v^2/2xx1/(2g)⇒h=v22×12g
h=v^2/(4g)h=v24g
Now |vecL|=hxxm|vecvcostheta|∣∣∣→L∣∣∣=h×m∣∣→vcosθ∣∣
Inserting calculated values we get
|vecL|=v^2/(4g)xxmxxv/sqrt2∣∣∣→L∣∣∣=v24g×m×v√2
=>|vecL|=(mv^3sqrt2)/(8 g)⇒∣∣∣→L∣∣∣=mv3√28g