A block of mass m is kept at the lowest point on a wedge of mass $M$ of height $h$ and it has an angle of inclination $\theta$. The wedge accelerates with an acceleration $a$ such that the mass $m$ is accelerated up the incline.
What will the trajectory of the block be with respect to ground when it has reached the highest point of the wedge and is just about to lose contact with it?
All the surfaces are smooth. Additional variables to solve this question may be defined if needed, as I myself don't know if additional info is required.
Hint.
Calling
$$ \cases{ F_a = m(a,0)\\ W=m(0,-g)\\ N=n(-\sin\theta,\cos\theta) } $$
we have
$$ F_a + W + N = m b(\cos\theta,\sin\theta) $$
where $b$ is the acceleration module along the ramp.
Now solving for $n,b$ we obtain
$$ b = a\cos\theta-g\sin\theta $$
so we need $a\cos\theta>g\sin\theta$
now along the ramp, the movement follows
$$ s = \frac 12 b t^2 $$