Eww imperial, but interesting maths nonetheless!

I think your calculations are slightly wrong because you are assuming that the skateboarder will come to a full stop at the end of the quarter-pipe, which is not true - they will keep most of their momentum, it will just be redirected.

Here’s my take on it

The world-record quarter-pipe drop (which happened on September 25th, and I think this comic was inspired by it) was on a 60 m high ramp. The skater experienced ~28 m/s and ~4 G. The ramp was far from optimal, with the most curvature at the bottom, so the tightest radius was actually less than 60 m. By a = v² / r ⇒ r = v² / a, we can estimate that the radius of the ramp at the most curved point was actually closer to 20 m (which about lines up with the images online). This doesn’t take into account acceleration due to gravity, so let’s be conservative and say 30 m, which is quite reassuring for us already, since we have a whooping 180 m to work with.

Now let’s calculate what the average acceleration would be throughout our trick in question, assuming a roughly constant speed over the entire quarter-pipe and only the centripetal force acting on the skateboarder. Of course neither will due to air resistance slowing the skateboarder down, and gravitational force increasing the contact force as time goes on, but those two are counteracting. Contact force increasing with gravity should win out (because air resistance depends on the velocity and so we get exponential decay or so, while contact force due to gravity will increase sinusoidaly or so), but I don’t think by much.

Assuming a circular quarter-pipe, the centripetal acceleration is |a| = v² / r. With a terminal velocity of ~50 m/s and a radius of 600 ft ≈ 180 m, we get |a| ≈ 14 m/s², or just 1.5 g (at least at the top of the quarter-pipe), which is really easy to pull off. Even if cueball wouldn’t slow down at all, that would mean 2.5 g at the bottom, which is still doable. Adding a perpendicular air resistance acceleration (which at least at the top will be equal to 1 g for obvious reasons) to both, we get a range from 1.8 g to 2.7 g, which is much less than the world record and so I deem doable.

Now let’s try being more precise.

Since the contact force is always perpendicular to the velocity vector, it will not change the speed; so, let’s try thinking in the reference frame of the rotating object.

So, the only changes in speed will come from the relationship between drag and gravity.

Let’s assume that acceleration due to drag is just C * |v|, i.e. some constant times the speed.

Let’s say x is the distance traveled along the ramp; then angle traveled α = x / (R * τ/4) radians.

In that case we get acceleration due to gravity = g (cos α) = g cos (x / (R * τ/4))

Finally,

x’’ = g (cos (x / (R * τ/4))) - C x’

where x₀ = 0, x’₀ = 50 m/s, R = 180 m, C = 1g / (50 m/s) ≈ 0.2 s⁻¹, g ≈ 10 m/s^2.

The units check out at least, so I think I’m mostly right.

I don’t know if there’s an analytical solution for this, but WolframAlpha refuses to find one.

Let’s try simplifying a bit by approximating the cos α as just 1 - 4α/τ (which is a bad approximation, but the best linear one, and the two-term Maclaurin series doesn’t lead to an analytic solution on Wolfram)

x’’ = g (1 - x / R) - C x’

This now does have a solution:

x(t) = 149.927 e^(-0.1 t) sin(0.213437 t) - 180 e^(-0.1 t) cos(0.213437 t) + 180

It is bad (doesn’t even reach 180 * τ/4, i.e. the skater doesn’t even reach the bottom of the quarter-pipe), so I don’t know how to proceed. Maybe what we need to take away from this is that air resistance is a big factor and the skateboarder will slow down very significantly.