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Fundamentals Of Physics
Found in: Page 143

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Short Answer

The terminal speed of a sky diver is 160 km/h in the spread-eagle position and 310 km/h in the nosedive position. Assuming that the diver’s drag coefficient C does not change from one position to the other, find the ratio of the effective cross-sectional area A in the slower position to that in the faster position.

The ratio of the effective cross-sectional area A in the slower position to that in the faster position is 3.75

See the step by step solution

Step by Step Solution

Step 1: Given

a=2mgCpv2t

which illustrates the inverse proportionality between the area and the speed-squared

thus

Aslow=310km/hAfast=160km/h

Step 2: Determining the concept

This problem is based on the drag force which is a type of friction. This is the force acting opposite to the relative motion of an object moving with respect to the surrounding medium. Using the concept of the drag force and terminal speed the ratio of the effective cross-sectional area A in the slower position to that in the faster position.

Formula:

The terminal speed is given by

vt=2FgCpA

Where C is the drag coefficient,p is the fluid density, A is the effective cross-sectional area, and Fg is the gravitational force.

solve for the area

A=2mgCpv2t

which illustrates the inverse proportionality between the area and the speed-squared

Step 3: Determining the ratio of the effective cross-sectional area A in the slower position to that in the faster position

When a ratio of areas has been setup of the slower case to the faster case,

AslowAfast=110km/h160km/h2 =3.75

Hence, the ratio of the effective cross-sectional area A in the slower position to that in the faster position is 3.75

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