QJEEM25Ph55a4: JEE Main 2025 Physics - Class 11 JEEM25Ph55a4 Solution: An object of mass ' m ' is projected from origin in a vertical…

$\vec{v_0} = \frac{v_0}{\sqrt{2}}\,\hat{i} + \frac{v_0}{\sqrt{2}}\,\hat{j}.$

At maximum height, the vertical component of the velocity becomes zero. Setting the vertical velocity to zero:

$\frac{v_0}{\sqrt{2}} - g\,t = 0 \quad \Longrightarrow \quad t = \frac{v_0}{g\sqrt{2}}.$

The coordinates at this time are:

$x = \frac{v_0}{\sqrt{2}}\,t = \frac{v_0}{\sqrt{2}} \cdot \frac{v_0}{g\sqrt{2}} = \frac{v_0^2}{2g},$

$y = \frac{v_0}{\sqrt{2}}\,t - \frac{1}{2}g\,t^2 = \frac{v_0^2}{2g} - \frac{1}{2}g\,\frac{v_0^2}{2g^2} = \frac{v_0^2}{2g} - \frac{v_0^2}{4g} = \frac{v_0^2}{4g}.$

Thus, the position vector at maximum height is:

$\vec{r} = \frac{v_0^2}{2g}\,\hat{i} + \frac{v_0^2}{4g}\,\hat{j}.$

At maximum height, the only nonzero component of velocity is the horizontal one:

$\vec{v} = \frac{v_0}{\sqrt{2}}\,\hat{i}.$

The angular momentum about the origin is given by:

$\vec{L} = m\,\vec{r} \times \vec{v}.$

Writing out the cross product:

$\vec{r} \times \vec{v} = \left(\frac{v_0^2}{2g}\,\hat{i} + \frac{v_0^2}{4g}\,\hat{j}\right) \times \frac{v_0}{\sqrt{2}}\,\hat{i}.$

Since the cross product of $\hat{i}$ with itself is zero and:

$\hat{j} \times \hat{i} = -\hat{k},$

we have:

$\vec{r} \times \vec{v} = \frac{v_0^2}{4g} \cdot \frac{v_0}{\sqrt{2}}\, (-\hat{k}) = -\frac{v_0^3}{4\sqrt{2}g}\,\hat{k}.$

Thus, the angular momentum is:

$\vec{L} = -\frac{m\,v_0^3}{4\sqrt{2}g}\,\hat{k}.$

This means the magnitude is:

$\frac{m\,v_0^3}{4\sqrt{2}g},$

and its direction is along the negative $\hat{k}$ (or negative $z$ -axis).