A stone is tied to a string of length L and i | vedclass

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(D) Let the lowest position be the origin $(0,0)$. The velocity at the lowest position is $\vec{v}_i = u\hat{i}$.When the string is horizontal, the st

Vedclass · Accepted Answer

$\sqrt{2(u^{2}-gL)}$

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(D) Let the lowest position be the origin $(0,0)$. The velocity at the lowest position is $\vec{v}_i = u\hat{i}$.When the string is horizontal, the stone has moved up by a height $h = L$. Using the law of conservation of energy:$\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgL$$v^2 = u^2 - 2gL$$v = \sqrt{u^2 - 2gL}$.At the horizontal position, the velocity vector is directed vertically upwards: $\vec{v}_f = \sqrt{u^2 - 2gL}\hat{j}$.The change in velocity is $\Delta\vec{v} = \vec{v}_f - \vec{v}_i = \sqrt{u^2 - 2gL}\hat{j} - u\hat{i}$.The magnitude of the change in velocity is $|\Delta\vec{v}| = \sqrt{(\sqrt{u^2 - 2gL})^2 + (-u)^2} = \sqrt{u^2 - 2gL + u^2} = \sqrt{2u^2 - 2gL} = \sqrt{2(u^2 - gL)}$.

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