QQuestionPhysics
QuestionPhysics
A skydiver weighing 200 lbs with clothes that have a drag coefficient of 0.325 is falling in an area with an atmospheric density of 1.225 kg/m² (assuming that altitude has a negligible effect on atmospheric density).
The skydiver can change body orientation from:
Straight-erect with a cross-sectional area of 6 sq ft.
Belly-flat with a cross-sectional area of 24 sq ft.
Calculate the terminal velocity of the person in both orientations:
A. Straight-erect
B. Belly-flat
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Answer
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Step 1:I'll solve this problem step by step, carefully following the LaTeX formatting guidelines:
Step 2:: Identify the Terminal Velocity Formula
- $$C_d$$ = drag coefficient
The terminal velocity is calculated using the equation: Where:
Step 3:: Convert Units
Belly-flat: $$24 \text{ sq ft} \times 0.0929 \frac{m^{2}}{\text{sq ft}} = 2.2296 \text{ m}^{2}
- Area conversion: A. B.
Step 4:: Calculate Terminal Velocity for Straight-Erect Position
v_{terminal} = 56.03 \frac{m}{s}
Step 5:: Calculate Terminal Velocity for Belly-Flat Position
v_{terminal} = 35.36 \frac{m}{s}
Final Answer
A. Straight-erect terminal velocity: 56.03 \frac{m}{s} B. Belly-flat terminal velocity: 35.36 \frac{m}{s} Key Insights: - Increasing cross-sectional area reduces terminal velocity - Drag coefficient and body orientation significantly impact terminal velocity - The skydiver falls much slower in the belly-flat position due to increased air resistance
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