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(D) The initial velocity components are $v_x = v_0 \cos 45^{\circ} = \frac{v_0}{\sqrt{2}}$ and $v_y = v_0 \sin 45^{\circ} = \frac{v_0}{\sqrt{2}}$.At m

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$\frac{m v_0^3}{4 \sqrt{2} g}$ along negative $z$-axis

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(D) The initial velocity components are $v_x = v_0 \cos 45^{\circ} = \frac{v_0}{\sqrt{2}}$ and $v_y = v_0 \sin 45^{\circ} = \frac{v_0}{\sqrt{2}}$.At maximum height $H$,the vertical velocity component is zero,and the horizontal velocity is $v_x = \frac{v_0}{\sqrt{2}}$.The maximum height $H$ is given by $H = \frac{v_y^2}{2g} = \frac{(v_0/\sqrt{2})^2}{2g} = \frac{v_0^2}{4g}$.The angular momentum $\vec{L}$ with respect to the origin is $\vec{L} = \vec{r} 	imes \vec{p} = m(\vec{r} 	imes \vec{v})$.At maximum height,the position vector is $\vec{r} = x\hat{i} + H\hat{j}$ and the velocity vector is $\vec{v} = v_x\hat{i}$.Thus,$\vec{L} = m(x\hat{i} + H\hat{j}) 	imes (v_x\hat{i}) = mH v_x (\hat{j} 	imes \hat{i}) = -mH v_x \hat{k}$.Substituting the values: $L = m \left( \frac{v_0^2}{4g} ight) \left( \frac{v_0}{\sqrt{2}} ight) = \frac{m v_0^3}{4 \sqrt{2} g}$.The direction is along the negative $z$-axis $(- \hat{k})$.

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