Motion on an Accelerating Incline
Let us analyze the motion in the non-inertial frame of reference of the wedge. The wedge is accelerating vertically upwards with acceleration $A$.
In this frame, there is a downward pseudo-force acting on the block equal to $mA$. This effectively increases the gravitational pull. The effective gravity in the wedge’s frame is:
$$g_{eff} = g + A$$
Within the wedge’s frame, the block simply slides down a frictionless incline under effective gravity $g_{eff}$. The acceleration of the block relative to the wedge, $a_{rel}$, is directed down the incline:
$$a_{rel} = g_{eff} \sin\theta = (g + A)\sin\theta$$
We are given that the block remains at a “level height” in the ground frame. This implies its absolute vertical acceleration is zero ($a_{y, abs} = 0$).
The absolute vertical acceleration is the sum of the wedge’s vertical acceleration and the vertical component of the block’s relative acceleration:
$$a_{y, abs} = A_{wedge} + (a_{rel})_y$$
The relative acceleration vector points down the incline (angle $\theta$ below horizontal). Its vertical component is downward (negative):
$$(a_{rel})_y = -a_{rel} \sin\theta = -(g + A)\sin\theta \cdot \sin\theta$$
$$a_{y, abs} = A – (g + A)\sin^2\theta$$
Setting the absolute vertical acceleration to zero:
$$A – (g + A)\sin^2\theta = 0$$
$$A = (g + A)\sin^2\theta$$
$$A = g\sin^2\theta + A\sin^2\theta$$
$$A(1 – \sin^2\theta) = g\sin^2\theta$$
$$A \cos^2\theta = g \sin^2\theta$$
$$A = g \frac{\sin^2\theta}{\cos^2\theta} = g \tan^2\theta$$