During descents, the negative slope of the hill in the power equation reflects the addition of gravitational potential energy to the power generated by the cyclist. In a freewheel (passive) descent, the cyclist's speed will be determined by the balance of the air resistance force and the gravitational force. As the cyclist accelerates, sv2 increases. Once kaAsv2 (plus the negligible power term associated with rolling resistance) increases to match giMs, the cyclist will reach terminal velocity. Any further increase in speed must be achieved by adding energy through pedaling. However, on steep hills, terminal velocities may reach 70 km·hr-1. At such high associated values of sv2, even the application of VO2max would result in only a minimal increase in speed.

Terminal velocity can be solved for in the cycling equation above by setting power at 0. If one assumes the rolling resistance term is also 0, and that there is no wind blowing (v = s), then the equation becomes:

kaAs3 = -giMs
or s = (-giM/kaA)1/2

Thus, the terminal velocity is roughly proportional to the square root of the ratio of M/A. Scaling reveals that larger cyclists have a greater ratio of mass to frontal area. They therefore descend hills faster as a consequence of purely physical, not physiological, laws. Since the larger cyclist has a greater mass, gravity acts on him or her with a greater force than it does on a smaller cyclist. (Note: A common misconception is to note the equal acceleration of two different sized objects in free fall in a vacuum, and assume that the force of gravity on both is equal. The force on the more massive object is greater, being exactly proportional to mass, which is why the more massive object is accelerated at the same rate as the less massive one.) While the larger cyclist also has a greater absolute frontal area than the smaller cyclist, the difference is not as great as that for their masses. Thus, the larger cyclist will attain a greater s3 before a balance of forces results in terminal velocity.