Would the terminal velocity of a roller coaster train differ on the track from if it was free falling through air? I'm working on a video about roller coaster science and I'm currently trying to figure out an average roller coasters terminal velocity going down a drop. All I can really do is fill in the information for terminal velocity in a free fall, but is that different from if it was going down the track? I assume it would add a slight drag. If so would it be that big of a difference or roughly the same? Lets assume this drop is going straight down 90 degrees as well.
 A: Terminal velocity is the velocity at which the force air resistance is equal and opposite to gravity in the direction of motion (ignoring friction for the moment). In free fall, the force of gravity is $F_g=mg$ where $m$ is the mass of the object in question, and $g$ is the gravitational acceleration at Earth's surface. On an incline, however, the force of gravity (in the direction of motion) is $F=mg\sin\theta$ where $\theta$ is the angle the incline makes with the horizontal. This should make sense because if $\theta=90^\circ$, then the incline is vertical, and the object is in free fall. If $\theta=0^\circ$, then the incline is actually a flat surface, and there is no motion.
I'm assuming you are working with a quadratic drag model where the drag force is given by $F_d=cv^2$ where $c$ is a constant having to do with the size and shape of the object, as well as the air density, and $v$ is velocity. Then if the free fall terminal velocity is $v_f$, $mg=cv_f^2$. On an incline, the terminal velocity is $v_i$, where $mg\sin\theta=cv_i^2$. Then substituting $cv_f^2$ for $mg$, $cv_f^2\sin\theta=v_i^2$ and $v_f\sqrt{\sin\theta}=v_i$. Thus, the incline terminal velocity and the free fall terminal velocity are related by the square root of the sine of the angle of incline.
All of this, of course, assumes that there is no friction. If we want to add a friction term, then we would balance forces to find that the force of friction is given by $F_f=\mu mg\cos\theta$, where $\mu$ is the coefficient of friction (approximately, there is a lot more going on in a complicated system like this, but this is as close as we can get without knowing specifics). So then the Terminal velocity is the number such that $mg\sin\theta-\mu mg\cos\theta-cv_i^2=0$. Doing the substitution again, we get $$v_f\sqrt{\sin\theta-\mu\cos\theta}=v_i$$
If, as you say, the incline is in fact $\theta=90^\circ$, then the roller coaster is in free fall unless the roller coaster is designed so that the wheels are always pressed against the track (and I'm not a roller coaster designer, but this seems like a good idea to me). The physics of this situation would depend on the design of the roller coaster.
A: Going down a track at an angle reduces the acceleration of gravity by the cosine of the angle.  If you then ignore all friction except air resistance (not a good assumption), the terminal velocity goes down as the square root of the acceleration of gravity.  If you want to get it right it won't be easy because there is a lot going on.  If you want it about right, this may help.
A: the terminal velocity would be the exact same no matter what incline be cause even though there is friction and an declined plane the terminal velocity is the same because friction just slows down the acceleration same with the declined plane the terminal velocity is when air resistance matches the amount of weight/force of the object   
