1. Interpret the inclinations

The planes make angles $30^\circ$ and $60^\circ$ with the vertical. So with the horizontal, their inclinations are:

• For block $A$: $60^\circ$
• For block $B$: $30^\circ$

2. Acceleration of each block along its plane

For a smooth fixed incline, acceleration along the plane is $a = g\sin\theta$ where $\theta$ is the angle with the horizontal.

Hence,

• For block $A$: $a_A = g\sin 60^\circ = \frac{\sqrt{3}}{2}g$

• For block $B$: $a_B = g\sin 30^\circ = \frac{1}{2}g$

3. Find vertical components of accelerations

We need the relative vertical acceleration, so only vertical components matter.

Since the acceleration is along the plane:

• Plane of $A$ is at $60^\circ$ to horizontal, so vertical component is $a_{Ay} = a_A \sin 60^\circ = \left(\frac{\sqrt{3}}{2}g\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{3}{4}g$ Direction is downward.

• Plane of $B$ is at $30^\circ$ to horizontal, so vertical component is $a_{By} = a_B \sin 30^\circ = \left(\frac{1}{2}g\right)\left(\frac{1}{2}\right)=\frac{1}{4}g$ Direction is downward.

4. Relative vertical acceleration of $A$ with respect to $B$

$a_{Ay} - a_{By} = \frac{3}{4}g - \frac{1}{4}g = \frac{1}{2}g$

Thus, $a_{A/B,\,vertical} = \frac{g}{2} = \frac{9.8}{2} = 4.9\,\text{m s}^{-2}$

Direction: downward for $A$ relative to $B$, i.e. along the vertical direction.

5. Check options

• A: $4.9\,\text{m s}^{-2}$ in horizontal direction $\rightarrow$ Incorrect
• B: $9.8\,\text{m s}^{-2}$ in vertical direction $\rightarrow$ Incorrect
• C: Zero $\rightarrow$ Incorrect
• D: $4.9\,\text{m s}^{-2}$ in vertical direction $\rightarrow$ Correct

Therefore, the correct option is D.