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(D) Using the law of conservation of energy between point $A$ and point $B$:$E_A = E_B$$\frac{1}{2}mv_A^2 + mgh = \frac{1}{2}mv_B^2$$v_B = \sqrt{v_A^2

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$3$

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(D) Using the law of conservation of energy between point $A$ and point $B$:$E_A = E_B$$\frac{1}{2}mv_A^2 + mgh = \frac{1}{2}mv_B^2$$v_B = \sqrt{v_A^2 + 2gh}$Given $v_A = 5\,m/s$,$g = 10\,m/s^2$,and $h = 10\,m$:$v_B = \sqrt{5^2 + 2 	imes 10 	imes 10} = \sqrt{25 + 200} = \sqrt{225} = 15\,m/s$At point $B$,the particle is at a distance $r = a = 10\,m$ from point $O$. The velocity vector at $B$ is perpendicular to the radius $OB$.The angular momentum $L$ about $O$ is given by:$L = m \cdot v_B \cdot r$$L = (20 	imes 10^{-3}\,kg) 	imes (15\,m/s) 	imes (10\,m)$$L = 0.02 	imes 150 = 3\,kg \cdot m^2/s$.

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