Air Resistance

In our work with trajectories, we ignored the friction of moving through the air. Air resistance increases as speed increases. The force of friction from air resistance is usually approximated as

F_air = Kv² = cAv²

where c is a constant that depends on the materials, and A is the cross sectional area of the falling body. Sometimes cA is replaced by a K. This means that the force due to air resistance is proportional to the square of the speed. When an object falls in air, the force from air resistance will increase as the object accelerates until it reaches a so-called "terminal" velocity. At this point, the force of air resistance and gravity equal each other. Thus the net force is 0, and there is no more acceleration. The falling object now has constant (terminal) velocity:

W = F_air

or:

m $\cdot$ g = cA $\cdot$ v²

Solving this for the terminal velocity we obtain:

\[ \rm v = \Bigg[{\frac{mg}{cA}\Bigg]}^{1/2} \]

Let's put in some numbers:

For an 80 kg skydiver falling with his body horizontal, cA is about 0.25 kg/m; so kg/m}\Bigg]}^{1/2} = 56 m/s \]
(which is about 126 miles/hour).

The demonstration video shows a simulation of the velocity of two dropping balls, one with and one without air resistance.