I would also like to know this.How do you actually calibrate the end speed by just watching a video? Can you please walk me through your calculations.

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Is your position mathematically supportable or is it merely an assumption based on unknowns?

I think he assumed constant acceleration, whereby an acceleration of 7.75m/s can be derived from the other two variables (time, distance). In absence of air resistance and friction, you would need 225kN of thrust to accelerate a 29t object at that rate.In the video 0:58, the J15 took off from extended position of 190m in 7 seconds, achieving approximate acceleration of 8m/s² and velocity of 200 kmph off the jump.

(Su33 stall speed is about 250kmph). Since the ski jump is angled at about 14°, Horizontal V is 190kmph and Vertical V is 50kmph. Vertical V is used as a safety against sink rate until target lift is achieved.

What I am skeptical is how the horizontal V was derived at the end of the ski-jump: I think the trajectory acceleration drops over the ski-jump section (due to impact and having to fight gravity).