Yes, the Hilltoppers were a popular quartet in the 50's. It always amazes me that my adult kids don't recognize names like this. They never saw Camelot, Carousel, or South Pacific. They may recognize Pat Boone's name, but can't name any of his songs. However, the 50's culture is not my topic. I want to discuss the scenario encountered by those who get their kicks topping a hill at high speed in an automobile.
In accident reconstruction involving hill topping, speed (V) can be determined by using a trajectory analysis of the vehicle during the time it is airborne. Knowing the takeoff position, [coordinates (X_{t},Y_{t})], the takeoff angle, and the landing position (Xl, Yl), V can be determined. The problem is that (X_{t},Y_{t}) is sometimes difficult to determine since there is no physical evidence to mark the position. However, analysis of the path can determine a relationship between speed and takeoff position.
When a car goes over a hill, the centrifugal force (C) acts in opposition to the normal component (N) of the force of gravity that holds the car on the road. If C = V^{2}•W/(g•R) is large enough, N becomes zero, and the car becomes airborne. W is the weight, g is the gravitational constant, and R is the radius of curvature of the path. Using Newton's laws, we see from the figure that during contact with the road:
(1) N=W•sinTV^{2}•W/(gR)
T is the angle of the velocity vector (tangent to the path) with respect to vertical. As N decreases with increasing speed, traction, the maximum braking force, and steering forces also decrease. Thus, the stopping distance increases and control decreases on a convex path.
When the car becomes airborne, N = 0, giving
(2) V^{2} /(gR) = sin T
This defines the relationship between takeoff speed and position along the path. The first point along the path with takeoff speed equal to or less than the speed of the car would be the takeoff point. If the path is a circular arc, the speed at which takeoff occurs is the greatest at the top of the hill (T=90°) where Vmax=(g•R)^{½}. The component of the force of gravity holding the car on the road is greatest at this point. At speeds greater than Vmax, equation 2 is invalid since the car is not in contact with the path. In this case, the car goes airborne at the first point on the circular path. This is a common occurrence.
The relationship between speed and takeoff position can be used in conjunction with the trajectory equations to determine the initial conditions for the airborne trajectory. The trajectory equations give:
(3) Xl =X_{t}+TV(sin T), and
(4) Yl =Y_{t}+TV(cos Tg)•T^{2}/2
T is the time between takeoff and impact. Unfortunately, an analytical solution requires the solution of transcendental equations. However, numerical solutions can be used. One approach is as follows:

Assume a value for V

Locate the first point on the path with takeoff velocity = V, and determine T, X_{t}, and Y_{t}

Determine T at impact from equation 3

Use T to determine Yl from equation 4

Compare the calculated value of Yl with the known impact point

Iterate values of V to satisfactory convergence

Using this method V can be found without prior knowledge of X_{t} and Y_{t}
For example, if a car traveling on a hill described by a circular arc with R = 299 ft goes airborne and lands at a point 100 ft horizontally and 17 ft vertically below the top of the hill, the speed is determined to be 66.7 mph.
Many things can happen while a car is airborne that can cause problems after landing. For example, it may be difficult to keep the wheels straight while airborne. If they are misaligned at impact, control may be lost. This leads to a whole other aspect of the investigation.