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Unformatted text preview: Answer, Key Homework 3 David McIntyre 45123 Mar 25, 2004 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page find all choices before making your selection. The due time is Central time. Chapters 2 and 3 problems. 001 (part 1 of 1) 10 points The height of a helicopter above the ground is given by h = ct 3 , where c = 1 . 1 m / s 3 , h is in meters, and t is in seconds. The helicopter takes off and after 4 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 . 9753 s. Explanation: Given : t = 4 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 4 s after takeoff: h = h ( t ) = (1 . 1 m / s 3 )(4 s) 3 = 70 . 4 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 . 1 m / s 3 )(4 s) 2 = 52 . 8 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 = (52 . 8 m / s) 2 + 2 ( 9 . 8 m / s 2 ) (70 . 4 m) = 4167 . 68 m 2 / s 2 , the time for the mailbag to reach the ground is t = v D g = 52 . 8 m / s p 4167 . 68 m 2 / s 2 9 . 8 m / s 2 = 11 . 9753 s . The negative solution is rejected. 002 (part 1 of 1) 0 points A freely falling body has a constant accelera tion of 9.8 m/s 2 . This means that: 1. the acceleration of the body decreases by 9.8 m/s 2 during each second. 2. the acceleration of the body increases by 9.8 m/s 2 during each second. 3. the body falls 9.8 m during each second. 4. the body falls 9.8 m during the first second only. 5. the speed of the body increases by 9.8 m/s during each second. correct Explanation: The acceleration is defined as the ratio of the change in velocity to the time period. If the acceleration is g = 9 . 8 m/s 2 , the change in velocity during one second is v = g t v = (9 . 8 m/s 2 )(1 s) = 9 . 8 m/s 003 (part 1 of 2) 5 points A stone is thrown upwards from the edge of a Answer, Key Homework 3 David McIntyre 45123 Mar 25, 2004 2 cliff y = 11 . 6 m high. It just misses the cliff on the way back down and hits the ground below, at y = 0, with a speed of v = 21 . 7 m / s. The acceleration of gravity is 9 . 8 m / s 2 . With what velocity was it released? Take up to be positive. Correct answer: 15 . 6054 m / s. Explanation: Solution: Since the acceleration is constant, v 2 = v 2 + 2 a ( y y ) v = 21 . 7 m / s , y = 11 . 6 m , and a = g , so v 2 = v 2 2 a ( y y ) = v 2 2( g )(0 y ) = ( 21 . 7 m / s) 2 2( 9 . 8 m / s 2 )( 11 . 6 m) = 243 . 53 m 2 / s 2 , so that v = 15 . 6054 m / s ....
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This note was uploaded on 09/15/2009 for the course PHY 557 taught by Professor Rijssenbeek during the Spring '08 term at Adelphi.
 Spring '08
 RIJSSENBEEK
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