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Unformatted text preview: Muraj, Hamza Homework 7 Due: Feb 6 2006, 4:00 am Inst: Florin 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The height of a helicopter above the ground is given by h = ct 3 , where c = 1 . 9 m / s 3 , h is in meters, and t is in seconds. The helicopter takes off and after 3 s it releases a small mailbag. The acceleration of gravity is 9 . 8 m / s 2 . How long after its release does the mailbag reach the ground? Correct answer: 11 . 3887 s. Explanation: Given : t = 3 s . Under free fall, h ( t ) = y ( t ) = y + v t + 1 2 at 2 . The initial height of the mailbag is the height of the helicopter 3 s after takeoff h = h ( t ) = (1 . 9 m / s 3 )(3 s) 3 = 51 . 3 m , and it starts its free fall motion from this point. Its initial velocity is equal to the veloc ity of the helicopter at that time v = dh dt = 3 ct 2 = 3(1 . 9 m / s 3 )(3 s) 2 = 51 . 3 m / s . Thus the equation of motion governing the mailbag is y ( t ) = 0 = h + v t 1 2 g t 2 . In quadratic form, 1 2 g t 2 v t h = 0 . From the quadratic formula, t = v q v 2 + 2 g h g . Since D = v 2 + 2 g h = (51 . 3 m / s) 2 + 2 ( 9 . 8 m / s 2 ) (51 . 3 m) = 3637 . 17 m 2 / s 2 , the time for the mailbag to reach the ground is t = v D g = 51 . 3 m / s p 3637 . 17 m 2 / s 2 9 . 8 m / s 2 = 11 . 3887 s . The negative solution is rejected. 002 (part 1 of 2) 10 points A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 34 . 6 below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline with a constant acceleration of 2 . 39 m / s 2 and travels 57 m to the edge of the cliff. The cliff is 33 . 3 m above the ocean....
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 Spring '09
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