A body of mass m is projected with speed v₀ at 45° to the horizontal. Its angular momentum about the point of projection at the highest point is:
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A body of mass m is projected with speed v₀ at 45° to the horizontal. Its angular momentum about the point of projection at the highest point is:

G1 Physics · Accepted Answer

At the highest point, horizontal speed is v₀/√2 and height H=v₀²/(4g). The perpendicular distance from the point of projection to the horizontal momentum line is H. Thus L=m(v₀/√2)H=mv₀³/(4√2g).