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Terminal velocity of a skydiver

Find the terminal velocity of an 85-kg skydiver falling in a spread-eagle position.

Strategy

At terminal velocity, F net = 0 . Thus, the drag force on the skydiver must equal the force of gravity (the person’s weight). Using the equation of drag force, we find m g = 1 2 ρ C A v 2 .

Solution

The terminal velocity v T can be written as

v T = 2 m g ρ C A = 2 ( 85 kg ) ( 9.80 m/s 2 ) ( 1.21 kg/m 3 ) ( 1.0 ) ( 0.70 m 2 ) = 44 m/s .

Significance

This result is consistent with the value for v T mentioned earlier. The 75-kg skydiver going feet first had a terminal velocity of v T = 98 m/s . He weighed less but had a smaller frontal area and so a smaller drag due to the air.

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Check Your Understanding Find the terminal velocity of a 50-kg skydiver falling in spread-eagle fashion.

34 m/s

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The size of the object that is falling through air presents another interesting application of air drag. If you fall from a 5-m-high branch of a tree, you will likely get hurt—possibly fracturing a bone. However, a small squirrel does this all the time, without getting hurt. You do not reach a terminal velocity in such a short distance, but the squirrel does.

The following interesting quote on animal size and terminal velocity is from a 1928 essay by a British biologist, J. B. S. Haldane, titled “On Being the Right Size.”

“To the mouse and any smaller animal, [gravity] presents practically no dangers. You can drop a mouse down a thousand-yard mine shaft; and, on arriving at the bottom, it gets a slight shock and walks away, provided that the ground is fairly soft. A rat is killed, a man is broken, and a horse splashes. For the resistance presented to movement by the air is proportional to the surface of the moving object. Divide an animal’s length, breadth, and height each by ten; its weight is reduced to a thousandth, but its surface only to a hundredth. So the resistance to falling in the case of the small animal is relatively ten times greater than the driving force.”

The above quadratic dependence of air drag upon velocity does not hold if the object is very small, is going very slow, or is in a denser medium than air. Then we find that the drag force is proportional just to the velocity. This relationship is given by Stokes’ law.

Stokes’ law

For a spherical object falling in a medium, the drag force is

F s = 6 π r η v ,

where r is the radius of the object, η is the viscosity of the fluid, and v is the object’s velocity.

Good examples of Stokes’ law are provided by microorganisms, pollen, and dust particles. Because each of these objects is so small, we find that many of these objects travel unaided only at a constant (terminal) velocity. Terminal velocities for bacteria (size about 1 μm ) can be about 2 μm/s . To move at a greater speed, many bacteria swim using flagella (organelles shaped like little tails) that are powered by little motors embedded in the cell.

Sediment in a lake can move at a greater terminal velocity (about 5 μm/s ) , so it can take days for it to reach the bottom of the lake after being deposited on the surface.

If we compare animals living on land with those in water, you can see how drag has influenced evolution. Fish, dolphins, and even massive whales are streamlined in shape to reduce drag forces. Birds are streamlined and migratory species that fly large distances often have particular features such as long necks. Flocks of birds fly in the shape of a spearhead as the flock forms a streamlined pattern ( [link] ). In humans, one important example of streamlining is the shape of sperm, which need to be efficient in their use of energy.

Questions & Answers

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The sonic range finder discussed in the preceding question often needs to be calibrated. During the calibration, the software asks for the room temperature. Why do you suppose the room temperature is required?
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Suppose a bat uses sound echoes to locate its insect prey, 3.00 m away. (See [link] .) (a) Calculate the echo times for temperatures of 5.00°C5.00°C and 35.0°C.35.0°C. (b) What percent uncertainty does this cause for the bat in locating the insect? (c) Discuss the significance of this uncertainty an
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The ear canal resonates like a tube closed at one end. (See [link]Figure 17_03_HumEar[/link].) If ear canals range in length from 1.80 to 2.60 cm in an average population, what is the range of fundamental resonant frequencies? Take air temperature to be 37.0°C,37.0°C, which is the same as body tempe
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Practice Key Terms 2

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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