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(C) The position vector of the particle at time $t$ is given by:$\vec{r} = (v_0 \cos 	heta) t \hat{i} + ((v_0 \sin 	heta) t - \frac{1}{2} g t^2) \ha

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$-\frac{1}{2} mg v_0 t^2 \cos 	heta \hat{k}$

Vedclass · Answer

(C) The position vector of the particle at time $t$ is given by:$\vec{r} = (v_0 \cos 	heta) t \hat{i} + ((v_0 \sin 	heta) t - \frac{1}{2} g t^2) \hat{j}$The velocity vector of the particle at time $t$ is given by:$\vec{v} = (v_0 \cos 	heta) \hat{i} + (v_0 \sin 	heta - gt) \hat{j}$The angular momentum $\vec{L}$ is defined as $\vec{L} = m(\vec{r} 	imes \vec{v})$:$\vec{L} = m [((v_0 \cos 	heta) t \hat{i} + ((v_0 \sin 	heta) t - \frac{1}{2} g t^2) \hat{j}) 	imes ((v_0 \cos 	heta) \hat{i} + (v_0 \sin 	heta - gt) \hat{j})]$Performing the cross product:$\vec{L} = m [((v_0 \cos 	heta) t) (v_0 \sin 	heta - gt) (\hat{i} 	imes \hat{j}) + ((v_0 \sin 	heta) t - \frac{1}{2} g t^2) (v_0 \cos 	heta) (\hat{j} 	imes \hat{i})]$Since $\hat{i} 	imes \hat{j} = \hat{k}$ and $\hat{j} 	imes \hat{i} = -\hat{k}$:$\vec{L} = m [((v_0^2 \sin 	heta \cos 	heta) t - v_0 g t^2 \cos 	heta) \hat{k} - ((v_0^2 \sin 	heta \cos 	heta) t - \frac{1}{2} v_0 g t^2 \cos 	heta) \hat{k}]$$\vec{L} = m [v_0^2 \sin 	heta \cos 	heta t - v_0 g t^2 \cos 	heta - v_0^2 \sin 	heta \cos 	heta t + \frac{1}{2} v_0 g t^2 \cos 	heta] \hat{k}$$\vec{L} = m [-\frac{1}{2} v_0 g t^2 \cos 	heta] \hat{k} = -\frac{1}{2} mg v_0 t^2 \cos 	heta \hat{k}$

$A$ small particle of mass $m$ is projected at an angle $\theta$ with the $x$-axis with an initial velocity $v_0$ in the $x-y$ plane as shown in the figure. At a time $t < \frac{v_0 \sin \theta}{g}$,the angular momentum of the particle is

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