An object of mass $$m$$ is projected from origin in a vertical  $$xy$$ plane at an angle $$45^\circ$$ with the $$x$$ -axis with an initial velocity $$v_0.$$ 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]

An object of mass $$m$$ is projected from the origin in the vertical xy-plane at an angle of $$45°$$ with initial speed $$v_0$$. To find its angular momentum about the origin at its maximum height, we first determine the time and position at that instant.

The time to reach the maximum height occurs when the vertical component of velocity becomes zero, namely at $$t = \frac{v_0 \sin 45°}{g} = \frac{v_0}{\sqrt{2}g}\,. $$ During this interval the horizontal component of motion remains uniform, so the x-coordinate is $$x = v_0 \cos 45° \times t = \frac{v_0}{\sqrt{2}} \times \frac{v_0}{\sqrt{2}g} = \frac{v_0^2}{2g}\,. $$ The maximum height itself is given by $$y = H = \frac{v_0^2 \sin^2 45°}{2g} = \frac{v_0^2}{4g}\,. $$ Hence the position vector at maximum height is $$\vec{r} = \frac{v_0^2}{2g}\,\hat{i} + \frac{v_0^2}{4g}\,\hat{j}\,. $$

At this instant the vertical velocity vanishes and only the horizontal component remains, so $$\vec{v} = v_0 \cos 45°\,\hat{i} = \frac{v_0}{\sqrt{2}}\,\hat{i}\,. $$

The angular momentum about the origin is given by $$\vec{L} = m\,\bigl(\vec{r}\times\vec{v}\bigr)\,. $$ Evaluating the cross product, $$\vec{r}\times\vec{v} = \Bigl(\frac{v_0^2}{2g}\hat{i} + \frac{v_0^2}{4g}\hat{j}\Bigr) \times \Bigl(\frac{v_0}{\sqrt{2}}\hat{i}\Bigr)\,. $$ Since $$\hat{i}\times\hat{i}=0$$ and $$\hat{j}\times\hat{i}=-\hat{k}$$, it follows that $$\vec{r}\times\vec{v} = \frac{v_0^2}{2g}\frac{v_0}{\sqrt{2}}\bigl(\hat{i}\times\hat{i}\bigr) + \frac{v_0^2}{4g}\frac{v_0}{\sqrt{2}}\bigl(\hat{j}\times\hat{i}\bigr) = -\frac{v_0^3}{4\sqrt{2}g}\,\hat{k}\,. $$ Therefore $$\vec{L} = m\bigl(\vec{r}\times\vec{v}\bigr) = -\frac{m\,v_0^3}{4\sqrt{2}g}\,\hat{k}\,. $$

The magnitude of the angular momentum is $$\frac{m\,v_0^3}{4\sqrt{2}g}$$ directed along the negative z-axis. The correct answer is Option 3: $$\frac{m\,v_0^3}{4\sqrt{2}g}$$ along the negative z-axis.