You have to clarify the question: dropping usually implies v = 0 initially, but you say v initial is 6 m/s and indicate that the ball is coming back to its original position (therefore, shot straight up into the air). The starting point is some position on the x-y plane (0,5 m) . By the "start" of the coordinate system, I imagine that you mean the origin (0,0). Clearly there is no angular momentum change if the ball just goes up in z directly above its starting position, because the velocity is parallel with the position vector: L = mv x r where v and r are vectors and x iindicates cross product or L = mvr sinθ which is equal to o for θ=0 Also, you need a mass in order to calculate the momentum.
OK, so let's solve for the angular momentum change as a function of time for r0 = (0,5,0)m, v0 = (0,0,6) m/s and a = (0,0,-9.80) m/s2
We can use the fact that rsinθ is the r perpendicular to the velocity in the z direction. (The projection of the position vector onto the xy plane) which is always 5 m from the origin.
L = mvrsinθ = 5mv v(t) = 6 m/s - (9.8 m/s2) t (Eqn. of motion)
L = 5m(6 - 9.8t) with ΔL = L(t) - L0 - 5m(6 - 9.8t) - 5m(6) = -5m(9.8t)
which is what you would get from Torgue (t) = ΔL The angular analog of Impulse Equation
because here the torque with respect to the origin is rFsinθ or -mg(5)
I hope that helped.
