Problem 1.14 - Problem 1.14 [Difficulty: 4] 2 Given: M = 70...

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Problem 1.14 [Difficulty: 4] Given: Data on sky diver: M7 0 k g = k 0.25 Ns 2 m 2 = Find: Maximum speed; speed after 100 m; plot speed as function of time and distance. Solution: Use given data; integrate equation of motion by separating variables. Treat the sky diver as a system; apply Newton's 2nd law: Newton's 2nd law for the sky diver (mass M) is (ignoring buoyancy effects): M dV dt Mg kV 2 = (1) Mg F D = kV 2 a= dV/dt (a) For terminal speed V t , acceleration is zero, so 2 0 = so V t k = V t 70 kg 9.81 × m s 2 m 2 0.25 N s 2 × 2 kg m × 1 2 = V t 52.4 m s = (b) For V at y = 100 m we need to find V ( y ). From (1) M dV dt M dV dy dy dt = MV dV dt = 2 = Separating variables and integrating: 0 V V V 1 2 d 0 y y g d = so ln 1 2 2k M y = or V 2 k 1e y M = Hence Vy () V t y M 1 2 = For y = 100 m: V 100 m ( ) 52.4 m s 2 0.25 × 2 m 2 100 × m 1 70 kg × kg m s 2 N × 1 2 = V 100 m
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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Problem 1.14 - Problem 1.14 [Difficulty: 4] 2 Given: M = 70...

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