Hm.. interesting

Thanks for the picture!

But another thing to note, it needs a constant acceleration for the plate to remain horizontal after getting there. So it's always acceleration. But of course, it might be like 1 degree lower than horizontal and from that point, it's not horizontal, though practically, it is. So, it'd be better if a certain range was assumed where the plate is considered horizontal.

I'll use base units to make it simpler for me.

The mass becomes 1.13 kg

The area becomes 0.5776 m^2

Also, I think that the height of the plate is required too, so that its centre of gravity can be estimated.

From that, assuming that the air remains still (no breeze from underneath, besides the one created by the motion of the vehicle), I can estimate the force that the air exerts on the plate, using

F = dm/dt * v

F = d(density*volume)/dt * v

F = d(density*length*area)/dt * v

F = density*area*v^2

Taking air density as 1.225 kg/m^3, the area as 0.5776, the force becomes F = 1.225Av^2

The area now varies with the angle which the plate makes with the horizontal, by the equation A = Ao cos(x) where Ao = 0.5776 m^2 and x is the angle.

F = 1.225(0.5776)cos(x) v^2

F = 0.7076v^2 cos(x)

The force that the plate exerts against the breeze is also dependent on the angle;

Fg = mg tan(x) = 11.058 tan(x)

So when the angle is 90 degrees with the vertical, and combining the two equations, we should get the speed required, but as I said earlier, that would be impossible. Let's take 80 degrees with the vertical to consider it horizontal.

F = 0.7076v^2 cos(x)

Fg = 11.058 tan(x)

0.7076v^2 cos(80) = 11.058 tan(80)

0.7076v^2 * 0.174 = 11.058*5.671

we get v = 509 m/s

or about 1100 mile/hr

That's huge, lol

Of course, acceleration now would put another variable into the game