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(A) The initial velocity components are $u_x = u \cos 45^{\circ} = 20 \sqrt{2} 	imes \frac{1}{\sqrt{2}} = 20 \,m/s$ and $u_y = u \sin 45^{\circ} = 20

Vedclass · Accepted Answer

$-400 \hat{k}$

Vedclass · Answer

(A) The initial velocity components are $u_x = u \cos 45^{\circ} = 20 \sqrt{2} 	imes \frac{1}{\sqrt{2}} = 20 \,m/s$ and $u_y = u \sin 45^{\circ} = 20 \sqrt{2} 	imes \frac{1}{\sqrt{2}} = 20 \,m/s$.After $t = 2 \,s$,the position coordinates are:$x = u_x t = 20 	imes 2 = 40 \,m$$y = u_y t - \frac{1}{2} g t^2 = 20 	imes 2 - \frac{1}{2} 	imes 10 	imes 2^2 = 40 - 20 = 20 \,m$.The velocity components at $t = 2 \,s$ are:$v_x = u_x = 20 \,m/s$$v_y = u_y - gt = 20 - 10 	imes 2 = 0 \,m/s$.The angular momentum $\vec{L}$ about the origin is given by $\vec{L} = \vec{r} 	imes \vec{p} = m(\vec{r} 	imes \vec{v})$.$\vec{r} = 40 \hat{i} + 20 \hat{j}$ and $\vec{v} = 20 \hat{i} + 0 \hat{j}$.$\vec{L} = 1 	imes [(40 \hat{i} + 20 \hat{j}) 	imes (20 \hat{i})] = 1 	imes [40 	imes 20 (\hat{i} 	imes \hat{i}) + 20 	imes 20 (\hat{j} 	imes \hat{i})]$.Since $\hat{i} 	imes \hat{i} = 0$ and $\hat{j} 	imes \hat{i} = -\hat{k}$,we get:$\vec{L} = 400(-\hat{k}) = -400 \hat{k} \,kg \cdot m^2/s$.

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