parachute_problem_extra

parachute_problem_extra - drag parachute 3 2 = = = ρ y D g...

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A person with a mass ( m ) of 115.5 kg (including all equipment) jumps from an airplane flying at an elevation of 4500 meters. The person falls, accelerating due to gravity until their velocity ( V ) reaches 53.6 m/s, at which point they open their parachute (Figure 1). Figure 1 – Diagram of deployed parachute and person for this exercise. a) Find the terminal velocity of the falling person once their parachute is fully deployed. Use the following constants and the principles of Newton’s second law of motion (See Figure 2 for a free body diagram, FDB.) , m kg 9093 . 0 density, air , s m 8 . 9 on, accelerati nal gravitatio , 45 . 1 t, coefficien
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Unformatted text preview: drag parachute 3 2 = = = ρ y D g C and, ( 29 , 2 1 , , 2 V AC D mg w a m F D f y y y = = = ↓ + ∑ where, ( 29 . s m velocity , s m on accelerati direction , m area exposed (N), force drag , (N) weight 2 2 = = = = = V y a A D w y f Figure 2 – Free body diagram (FBD) diagram for this exercise. Figure 3 – Diagram of parachute exposed area with diameter dimensions shown. b) Find the time (sec) it takes for the person to accelerate to the terminal velocity after their parachute has fully opened. Hint: Recall that dt V d a =...
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This note was uploaded on 02/18/2012 for the course EML 3035 taught by Professor Daly,j during the Spring '08 term at University of South Florida.

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parachute_problem_extra - drag parachute 3 2 = = = ρ y D g...

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