Two fixed frictionless inclined plane making an angle ${30}^o$ and ${60}^o$ with the vertical are shown in the figure. Two blocks $A$ and $B$ are placed on the two planes. What is the relative vertical acceleration of $A$ with respect to $B$?

The correct option is D $4.9m{s}^{-2}$ in vertical direction

For the motion of the block along the inclined plane,
$mg\sin \theta =m{a}_c$

$a_c=g\sin \theta$

Where ‘$a_c$’ is the acceleration along the inclined plane.

On resolving $a_c$ in vertical and horizontal components, from the figure, we can see that the vertical component is
$a_{cv}=g{\sin }^2\theta$

Now for block A, vertical acceleration is
$a_A=g{\sin }^{2}60$
Now for block B, vertical acceleration is
$a_B=g{\sin }^{2}30$

$\therefore$ The relative vertical acceleration of A w.r.t B is

$a_{AB}=g({\sin }^{2}60-{\sin }^{2}30)=g(\frac{3}{4}-\frac{1}{4})=\frac{g}{2}=4.9m/s^2$ in the vertical direction

Hence, the answer is OPTION D.