Question 198 of 280

An object of mass ' m ' is projected from origin in a vertical $x y$ plane at an angle $45^{\circ}$ with the x axis with an initial velocity $\mathrm{v}_0$. The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [ g is acceleration due to gravity]

A$\frac{m v_o^3}{2 \sqrt{2} g}$ along negative $z$-axisB$\frac{m v_o^3}{4 \sqrt{2} g}$ along positive $z$-axisC$\frac{m v_o^3}{4 \sqrt{2} g}$ along negative $z$-axisD$\frac{m v_o^3}{2 \sqrt{2} g}$ along positive $z$-axis

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Answer: (C) $\frac{m v_o^3}{4 \sqrt{2} g}$ along negative $z$-axis

$\begin{aligned}
& \mathrm{H}=\frac{\left(\frac{\mathrm{v}_0}{\sqrt{2}}\right)^2}{2 \mathrm{~g}}=\frac{\mathrm{v}_0^2}{4 \mathrm{~g}} \\
& \mathrm{~L}=\mathrm{mvh} \\
& \mathrm{~L}=\mathrm{m} \frac{\mathrm{v}_0}{\sqrt{2}} \frac{\mathrm{v}_0^2}{4 \mathrm{~g}}
\end{aligned}$