A small particle of mass m is projected at an angle $θ$ with the x-axis with an initial velocity us in the x-y plane as shown in the figure. At a time $t < \frac{v_{0} \sin θ}{g}$ , angular momentum of the particle is :

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$- \frac{1}{2} {mgv}_{0} t^{2} \cos θ \hat{k}$

${mgv}_{0} tcos θ \hat{k}$

$\frac{1}{2} {mgv}_{0} t^{2} \cos θ \hat{i}$

$- {mgv}_{0} t^{2} \cdot \cos θ \hat{j}$

answer is D.

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Detailed Solution

$\begin{array}{l} \vec{L} = \vec{r} \times \vec{p} where \vec{r} = x \hat{i} + y \hat{j} = v_{0} \cos θt \hat{i} + (v_{0} \sin θt - \frac{1}{2} {gt}^{2}) \hat{j} \\ and \vec{p} = m \vec{v} = m [v_{0} \cos θ \hat{i} + (v_{0} \sin θ - gt) \hat{j}] \\ So, \vec{j} = |\begin{array}{ccc} \hat{i} & \hat{j} \\ v_{0} \cos θt & v_{0} \sin θt - \frac{1}{2} {gt}^{2} & 0 \\ {mv}_{0} \cos θ & m (v_{0} \sin θ - gt) & 0 \end{array}| = - \frac{1}{2} {mgv}_{0} t^{2} \cos θ \hat{j} \end{array}$