Relative motion

Let's invoke the Virtual Work done method!!

This method states that the work done by the internal forces in case of a mechanical system is always zero. In this case, the internal forces are the tension in the strings. Let string attached with block $'A'$ has a tension $'T'$ in it. Clearly, the string attached to block $'B'$ has tension $'2T'$ in it.

Now, the velocity component of $'A'$ in the direction of tension is ${ v }_{ A }\cos { \theta }$ . So, by virtual work done:

${ T }_{ A }{ x }_{ A }+{ T }_{ B }{ x }_{ B }=\quad 0$

Differentiating this equation with respect to time, we can get that:

${ T }_{ A }{ v }_{ A }+{ T }_{ B }{ v }_{ B }=\quad 0$ $T{ v }_{ A }\cos { \theta } +{ 2T }{ v }_{ B }\quad =\quad 0\\$

From the adjacent triangle,

$\cos { \theta \quad =\frac { base }{ hypotenuse } } \\ \quad =\frac { x }{ \sqrt { { x }^{ 2 }+{ h }^{ 2 } } }$

Using this information, and ignoring the negative sign, which gives direction, we get:

${ v }_{ B }\quad =\quad \frac { T{ v }_{ A }\cos { \theta } }{ 2T } \\ \quad \quad \quad =\quad \frac { x{ v }_{ A } }{ 2\sqrt { { x }^{ 2 }+{ h }^{ 2 } } }$

CHEERS!!:)

$\large Tan\theta=\dfrac h x,~~and~~Sin\theta=\dfrac h R, ~~where ~~R^2=h^2+x^2. \\\implies h=x*Tan\theta,~~and~~h=R*Sin\theta. ~~Cos\theta=\dfrac{x}{\sqrt{x^2+h^2} }\\Differantiating~w.r.t.~~time ~and~equating~\dfrac{d\theta}{dt} from~ both, \\\dfrac{V_A* Tan\theta}{x*sec^2 \theta}=\dfrac{V_R*Sin\theta}{R*Cos\theta.} ~~\therefore V_R=\dfrac{V_A* Tan\theta}{x*sec^2 \theta}*\dfrac{R*Cos\theta} {Sin\theta} .\\But~V_B=\dfrac 1 2 *V_R=\dfrac 1 2 *V_A*\dfrac{x}{\sqrt{x^2+h^2} } .$