An object of mass ' m ' is projected from origin in a vertical xy plane at an angle $45^{\circ}$ with the $\mathrm{x}-$ 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]

<p>$\vec{v_0} = \frac{v_0}{\sqrt{2}}\,\hat{i} + \frac{v_0}{\sqrt{2}}\,\hat{j}.$</p> <p>At maximum height, the vertical component of the velocity becomes zero. Setting the vertical velocity to zero:</p> <p>$$ \frac{v_0}{\sqrt{2}} - g\,t = 0 \quad \Longrightarrow \quad t = \frac{v_0}{g\sqrt{2}}. $$</p> <p>The coordinates at this time are:</p> <p>$$ x = \frac{v_0}{\sqrt{2}}\,t = \frac{v_0}{\sqrt{2}} \cdot \frac{v_0}{g\sqrt{2}} = \frac{v_0^2}{2g}, $$</p> <p>$$ y = \frac{v_0}{\sqrt{2}}\,t - \frac{1}{2}g\,t^2 = \frac{v_0^2}{2g} - \frac{1}{2}g\,\frac{v_0^2}{2g^2} = \frac{v_0^2}{2g} - \frac{v_0^2}{4g} = \frac{v_0^2}{4g}. $$</p> <p>Thus, the position vector at maximum height is:</p> <p>$\vec{r} = \frac{v_0^2}{2g}\,\hat{i} + \frac{v_0^2}{4g}\,\hat{j}.$</p> <p>At maximum height, the only nonzero component of velocity is the horizontal one:</p> <p>$\vec{v} = \frac{v_0}{\sqrt{2}}\,\hat{i}.$</p> <p>The angular momentum about the origin is given by:</p> <p>$\vec{L} = m\,\vec{r} \times \vec{v}.$</p> <p>Writing out the cross product:</p> <p>$$ \vec{r} \times \vec{v} = \left(\frac{v_0^2}{2g}\,\hat{i} + \frac{v_0^2}{4g}\,\hat{j}\right) \times \frac{v_0}{\sqrt{2}}\,\hat{i}. $$</p> <p>Since the cross product of $\hat{i}$ with itself is zero and:</p> <p>$\hat{j} \times \hat{i} = -\hat{k},$</p> <p>we have:</p> <p>$$ \vec{r} \times \vec{v} = \frac{v_0^2}{4g} \cdot \frac{v_0}{\sqrt{2}}\, (-\hat{k}) = -\frac{v_0^3}{4\sqrt{2}g}\,\hat{k}. $$</p> <p>Thus, the angular momentum is:</p> <p>$\vec{L} = -\frac{m\,v_0^3}{4\sqrt{2}g}\,\hat{k}.$</p> <p>This means the magnitude is:</p> <p>$\frac{m\,v_0^3}{4\sqrt{2}g},$</p> <p>and its direction is along the negative $\hat{k}$ (or negative $z$-axis).</p>