Today I was driving on a hill side and on the opposite lane this very careless gentleman was traveling at a very high speed on a large truck that I surely thought his truck would of lifted off the ground. I started to think about how one would even calculate at what speed a certain mass would need to be traveling in order to launch itself from a cliff that makes an angle $\theta$ and how far that object would land.
It seemed like a fairly easy problem at first, so I drew a free body diagram with all (to my knowledge) possible variables present and how chancing them would affect the results. However I am unsure where to start.
Best Answer
The car becomes "weightless" when the curvature of the road is sufficient that the car does not stay connected to the road. Angles don't really matter - what matters is a change in the direction of the road.
As you know, an object going around in a circle needs a force $F = \frac{mv^2}{r}$ to stay in the orbit at radius $r$. Normally, when you drive over a bump the force of gravity is sufficient to keep you connected - but now we can see that the car will lift off if the radius of curvature $r$ is too small:
$$\frac{mv^2}{r} \gt mg\\ r \lt \frac{v^2}{g}$$
For a car going 100 km/h, that puts the limiting radius at about 80 m. Another way of thinking about this: if you think of the car driving off a cliff, it would drop 5 m in the first second. If the road drops away faster than that, the car will appear to lift off. That's perhaps more intuitive than the radius of curvature.