The terminal speed of a sky diver is 157 km/h in the spread-eagle position and 330 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.
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To find the ratio of the effective cross-sectional area (A) in the slower position to that in the faster position, we can use the relationship between terminal speed and drag force.
The drag force (F) experienced by an object in a fluid is given by the equation:
F = (1/2) * ρ * v^2 * C * A,
Where:
- ρ is the density of the fluid (air, in this case),
- v is the velocity of the object relative to the fluid (terminal speed),
- C is the drag coefficient, which indicates the object's shape and surface properties,
- A is the effective cross-sectional area of the object.
Given that the terminal speed in the spread-eagle position is 157 km/h (or 43.6 m/s) and the terminal speed in the nosedive position is 330 km/h (or 91.7 m/s), we can set up the following equation:
(1/2) * ρ * (43.6^2) * C * A_slow = (1/2) * ρ * (91.7^2) * C * A_fast.
Since the densities and drag coefficients cancel out, we can simplify the equation to find the ratio of the effective cross-sectional areas:
A_slow / A_fast = (91.7^2) / (43.6^2).
Calculating the ratio:
A_slow / A_fast ≈ 3766.89 / 1900.96 ≈ 1.9806.
Therefore, the ratio of the effective cross-sectional area in the slower position to that in the faster position is approximately 1.9806.