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Question

(A) 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 \cdot t - \frac{1}{2} g

Vedclass · Accepted Answer

$-\frac{1}{2} mg v_{0} t^{2} \cos 	heta \hat{k}$

Vedclass · Answer

(A) The position vector of the particle at time $t$ is given by:$\vec{r} = (v_{0} \cos 	heta) t \hat{i} + (v_{0} \sin 	heta \cdot t - \frac{1}{2} g t^{2}) \hat{j}$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 $\vec{L}$ about the origin is given by $\vec{L} = \vec{r} 	imes \vec{p} = m(\vec{r} 	imes \vec{v})$.$\vec{L} = m [((v_{0} \cos 	heta) t \hat{i} + (v_{0} \sin 	heta \cdot t - \frac{1}{2} g t^{2}) \hat{j}) 	imes ((v_{0} \cos 	heta) \hat{i} + (v_{0} \sin 	heta - gt) \hat{j})]$Using the cross product rules $\hat{i} 	imes \hat{i} = 0$,$\hat{j} 	imes \hat{j} = 0$,$\hat{i} 	imes \hat{j} = \hat{k}$,and $\hat{j} 	imes \hat{i} = -\hat{k}$:$\vec{L} = m [((v_{0} \cos 	heta) t)(v_{0} \sin 	heta - gt) \hat{k} + (v_{0} \sin 	heta \cdot t - \frac{1}{2} g t^{2})(v_{0} \cos 	heta) (-\hat{k})]$$\vec{L} = m [v_{0}^{2} \sin 	heta \cos 	heta \cdot t - v_{0} g \cos 	heta \cdot t^{2} - v_{0}^{2} \sin 	heta \cos 	heta \cdot t + \frac{1}{2} g v_{0} \cos 	heta \cdot t^{2}] \hat{k}$$\vec{L} = m [-\frac{1}{2} g v_{0} \cos 	heta \cdot t^{2}] \hat{k} = -\frac{1}{2} mg v_{0} t^{2} \cos 	heta \hat{k}$

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