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

The planes are smooth, so each block accelerates only under the component of gravity that acts parallel to its incline.

Let $$\theta$$ be the angle an incline makes with the horizontal. The acceleration of a block sliding down a smooth incline is then
$$a = g\sin\theta$$

The statement “plane makes an angle $$\alpha$$ with the vertical” means it makes an angle $$90^\circ-\alpha$$ with the horizontal.

Case A (block $$A$$)

Given: incline makes $$30^\circ$$ with the vertical.
Hence $$\theta_A = 90^\circ-30^\circ = 60^\circ$$ with the horizontal.

Acceleration along the plane:
$$a_A = g\sin60^\circ = g\left(\frac{\sqrt3}{2}\right)$$

The direction of motion is along the plane, which is $$30^\circ$$ away from the vertical. Therefore the vertical component of $$a_A$$ is
$$a_{A,v} = a_A\cos30^\circ = g\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt3}{2}\right)=\frac{3g}{4}$$

Case B (block $$B$$)

Given: incline makes $$60^\circ$$ with the vertical.
Hence $$\theta_B = 90^\circ-60^\circ = 30^\circ$$ with the horizontal.

Acceleration along the plane:
$$a_B = g\sin30^\circ = g\left(\frac12\right)$$

The motion is $$60^\circ$$ away from the vertical, so its vertical component is
$$a_{B,v} = a_B\cos60^\circ = g\left(\frac12\right)\left(\frac12\right)=\frac{g}{4}$$

Relative vertical acceleration of $$A$$ with respect to $$B$$ (downward taken as positive):
$$a_{\text{rel}} = a_{A,v} - a_{B,v} = \frac{3g}{4}-\frac{g}{4}=\frac{g}{2}=4.9\ \text{m\,s}^{-2}$$

Therefore the relative acceleration is $$4.9\ \text{m\,s}^{-2}$$ in the vertical (downward) direction.

Option D which is: $$4.9$$ ms$$^{-2}$$ in vertical direction