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80 kg human falling from 10 m in atmosphere of 4200 millibar, affected by .5 G of gravity

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Question: How big shall the area of a parachute be to let a 80 kilogram human land in 10 metres/second in an atmosphere of 4200 millibar and affected by 0.5 G of gravity?

Asked by ahrvid

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To calculate the required parachute area, we'll need to use the following formula: Area = (2 * Mass * Speed) / (Air density * Drag coefficient * Gravity) First, let's break down and gather the known variables: Mass (m) = 80 kg Speed (v) = 10 m/s Gravity (g) = 0.5 * 9.81 m/s²; considering Earth's gravity = 4.905 m/s² Air pressure (P) = 4200 millibar (1 millibar = 100 Pascal); So, P = 4200*100 = 420,000 Pa Ideal gas constant (R) = 287 J/kg.K (for dry air) Temperature (T) = assuming standard room temperature = 20°C = 293.15 K Drag coefficient (Cd) = approximately 1.5 for a typical round parachute Now, we will first calculate the air density (ρ). The formula for air density (ρ) in terms of air pressure (P), the gas constant (R) and temperature (T) is: ρ = P / (R * T) ρ = 420,000 Pa / (287 J/kg.K * 293.15 K) ρ ≈ 5.009 kg/m³ Now, let's calculate the required parachute area (A): Area = (2 * 80 kg * 10 m/s) / (5.009 kg/m³ * 1.5 * 4.905 m/s²) Area ≈ 43.42 m² Therefore, the area of the parachute should be approximately 43.42 square meters to let an 80 kg human land at 10 m/s in the given atmospheric conditions.

Answered by gpt-4

April 25, 2023, 9:33 a.m.

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This article was generated in part by one of OpenAI’s large-scale language-generation model, answering user ahrvid's question. Images were generated with OpenAI's AI model, DALL·E 2. The Internet takes ultimate responsibility for the content of this publication.
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Published: Tuesday, April 25, 2023

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