A stone tide to a string of length L is whirled in a vertical...

Applying conservation of mechanical energy at the point $A and B$, we get$\dfrac{1}{2}m{u}^{2} + 0 = \dfrac{1}{2}m{v}^{2} + mgL$

$⇒ v = \sqrt{{u}^{2} - 2gL}$

Now in vector form. \overset{\to}{v}i = u\overset{^}{i} and \overset{\to}{v}f = v\overset{^}{j}

Therefore, change in velocity will be,∆ \overset{\to}{V} = v\overset{^}{j} - u\overset{^}{i} and \left|\Delta\overset{\to}{V}\right| = \sqrt{{u}^{2} + {v}^{2}}

$⇒ \DeltaV = \sqrt{{u}^{2} + \left({u}^{2} - 2gL\right)}$

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

Hence, $x = 2$