How do you know? If the gravitation is absent, how does the rod fall? You are grasping at straws, unwilling to admit that you are wrong again.
Prove it. Stop waving your arms, use math. While you are at it, see if you can also neglect the Thomas precession. This is a much tougher problem than the mockery you are trying to make it to be.
Gravity i snot the issue here. Neddy Bate is correct in his statement that the body cannot land flat in the embankment. We know that the body lans flat in the train frame. Thus if we were to put clocks on the floor of the train at the points where the head and feet land, and those clocks are synchronized in the train frame, they each will have identical readings at impact. But we also know that these clocks are in motion with respect to the embankment and thus are subject to the relativity of simultaneity and cannot be in sync according to the embankment frame. The reading on the clock where the head impacts and the reading on the clock where the feet impact is invariant between frames. Thus the embankment frame must agree that the two clocks read exactly the same at impact. And since the clocks cannot read exactly the same at a given instant in the embankment frame, the head and feet cannot land at the same instant in the embankment frame.
This is true whether the body was falling due to gravity or just traveling downward at a constant velocity. Gravity adds an extra complication if you want to do an in depth analysis, but the existence of gravity is not relevant to the solution of the main issue; If one end hits first in one frame, would the force distribution be different than that in the frame in which it lands flat.
The answer to this is in the difference between time-like and space-like intervals.
Consider the normal situation of a rod falling at an angle. When the first end hits the shock of impact travels through the rod effecting the rest of the rod. Thus if you pick some point along the rod, when the force caused by the impact of the end reaches it acts to counter the points downward momentum. in turn some of its momentum is transfered back down the rod to the impact point. If the force of this interaction is greater than the strength of the rod, the rod will suffer some permanent deformation. Also, since out point along the rod has given up some of its momentum, when it does finally impact it will be at a lesser speed than the end that hit first. The upshot is that the force of impact is not equally distributed along the length of the rod,
Since there is time enough between the initial end hitting and the point along the rod hitting the surface for there to be a causal link between them (the impact at the end has an effect on the impact of the point along the rod), there is a time-like interval between the two events.
Now let's consider the situation in the OP. here we hava an object that is falling skewed due to RoS while at same time having a relative sideways motion (in the embankment).
For our example, assume that the relative velocity is from left to right. This means the the left end will be lower and hit first. when it hits, the impact force travels through the rod. For simplicities sake, we'll say that it moves at the greatest possible speed, that of light. So, from the embankment frame, the impact moves through the rod at c from left to right. however, since the rod itself is also moving from left to right, the speed that the impact moves with respect to the rod is c-v. if x is the distance of some point of the rod form the end in the train frame, then the time it takes for the impact to propagate form the end to this point is
$$\frac{x}{\gamma (c-v)}$$
The time delay between the end and the same point is:
$$\frac{xv}{\gamma c^2}$$
For any value of x or v, the RoS time delay will be shorter than the time needed for the impact to propagate to the same point.
So what does this mean? It means that due to the RoS, the delay between the end and some point along the rod hitting the surface is less than the time it takes for the impulse of the end's impact to reach this point. Which means that every point along the rod's length hits the surface without being affected by the impact of the points of the rod that impacted before it (and vice versa) . Since there is no causal link between the impacts at different points of the rod, the interval between these events are space-like. This also means each point of the rod hits with the same force and speed as every other part of the rod, and thus the total force of impact is evenly distributed along the length of the rod, just like it it in the train frame.
