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(B) Let the lowest position be $A$ and the horizontal position be $B$. At $A$, the velocity is $\vec{v}_A = u \hat{i}$.Using the conservation of mecha

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$2$

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(B) Let the lowest position be $A$ and the horizontal position be $B$. At $A$, the velocity is $\vec{v}_A = u \hat{i}$.Using the conservation of mechanical energy between $A$ and $B$: $\frac{1}{2}mu^2 = \frac{1}{2}mv_B^2 + mgL$.Solving for $v_B$, we get $v_B^2 = u^2 - 2gL$, so $v_B = \sqrt{u^2 - 2gL}$.At position $B$, the velocity is $\vec{v}_B = v_B \hat{j} = \sqrt{u^2 - 2gL} \hat{j}$.The change in velocity is $\Delta \vec{v} = \vec{v}_B - \vec{v}_A = \sqrt{u^2 - 2gL} \hat{j} - u \hat{i}$.The magnitude of the change in velocity is $|\Delta \vec{v}| = \sqrt{(-u)^2 + (\sqrt{u^2 - 2gL})^2} = \sqrt{u^2 + u^2 - 2gL} = \sqrt{2u^2 - 2gL}$.$|\Delta \vec{v}| = \sqrt{2(u^2 - gL)}$.Comparing this with $\sqrt{x(u^2 - gL)}$, we find $x = 2$.

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