# 6.4 Drag force and terminal speed  (Page 2/12)

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Typical values of drag coefficient C
Object C
Airfoil 0.05
Toyota Camry 0.28
Ford Focus 0.32
Honda Civic 0.36
Ferrari Testarossa 0.37
Dodge Ram Pickup 0.43
Sphere 0.45
Hummer H2 SUV 0.64
Skydiver (feet first) 0.70
Bicycle 0.90
Skydiver (horizontal) 1.0
Circular flat plate 1.12

Substantial research is under way in the sporting world to minimize drag. The dimples on golf balls are being redesigned, as are the clothes that athletes wear. Bicycle racers and some swimmers and runners wear full bodysuits. Australian Cathy Freeman wore a full body suit in the 2000 Sydney Olympics and won a gold medal in the 400-m race. Many swimmers in the 2008 Beijing Olympics wore (Speedo) body suits; it might have made a difference in breaking many world records ( [link] ). Most elite swimmers (and cyclists) shave their body hair. Such innovations can have the effect of slicing away milliseconds in a race, sometimes making the difference between a gold and a silver medal. One consequence is that careful and precise guidelines must be continuously developed to maintain the integrity of the sport.

## Terminal velocity

Some interesting situations connected to Newton’s second law occur when considering the effects of drag forces upon a moving object. For instance, consider a skydiver falling through air under the influence of gravity. The two forces acting on him are the force of gravity and the drag force (ignoring the small buoyant force). The downward force of gravity remains constant regardless of the velocity at which the person is moving. However, as the person’s velocity increases, the magnitude of the drag force increases until the magnitude of the drag force is equal to the gravitational force, thus producing a net force of zero. A zero net force means that there is no acceleration, as shown by Newton’s second law. At this point, the person’s velocity remains constant and we say that the person has reached his terminal velocity     $\left({v}_{\text{T}}\right).$ Since ${F}_{\text{D}}$ is proportional to the speed squared, a heavier skydiver must go faster for ${F}_{\text{D}}$ to equal his weight. Let’s see how this works out more quantitatively.

At the terminal velocity,

${F}_{\text{net}}=mg-{F}_{\text{D}}=ma=0.$

Thus,

$mg={F}_{\text{D}}.$

Using the equation for drag force, we have

$mg=\frac{1}{2}C\rho A{v}_{\text{T}}^{2}.$

Solving for the velocity, we obtain

${v}_{\text{T}}=\sqrt{\frac{2mg}{\rho CA}}.$

Assume the density of air is $\rho =1.21\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}.$ A 75-kg skydiver descending head first has a cross-sectional area of approximately $A=0.18\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}$ and a drag coefficient of approximately $C=0.70$ . We find that

${v}_{\text{T}}=\sqrt{\frac{2\left(75\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(9.80\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)}{\left(1.21\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}\right)\left(0.70\right)\left(0.18\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}\right)}}=98\phantom{\rule{0.2em}{0ex}}\text{m/s}=350\phantom{\rule{0.2em}{0ex}}\text{km/h}\text{.}$

This means a skydiver with a mass of 75 kg achieves a terminal velocity of about 350 km/h while traveling in a pike (head first) position, minimizing the area and his drag. In a spread-eagle position, that terminal velocity may decrease to about 200 km/h as the area increases. This terminal velocity becomes much smaller after the parachute opens.

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