What would be the estimated terminal of the placard?
ok. That placard has no terminal v until it sheds its forward momentum and clears the 727 vortex wake behind the plane. That placard does not behave like the box dropped during the test flight. Remember, the plane is still moving ahead at 200mph even as the placard has separated from the plane. The hi velocity vortex from the plane and its engines will catch up with the placard which is now shedding velocity ... those forces could conceivably lift the placard up vertically at first and stall its descent, or make it travel in loops, or tear it apart ... until gravity becomes the dominant force on the placard.
I have always wondered if the placard looks like it does (half an original placard) because it was torn apart in the jet's vortex? May the other half is laying on the ground somewhere? That's a job for Kermit to find and solve!
This is not a single variable linear problem but a multi-variant problem for calculus! We have been trying to tell people that for a long time. Nobody is listening?
In the photo attached add another large vortex behind the plane (unseen in this photo) from the three engines all clustered. Those gases are propagating faster than any terminal v initially involved ... by a long long way! These vortexes behind the plane (as it is moving ahead) are propagating faster than any terminal-v the placard can ever have due to gravity! Now throw in winds and the direction of winds, once the placard manages to survive the vortexes and gets to less turbulent air which has no 'lift', so the plastic card (whats left of it) can begin to fall to have a terminal v. Then ask your question.
Im not saying anything that R99, 377, TK, Hominid, etal dont already know.
