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Unformatted text preview: Overton, Mays Homework 7 Due: Feb 8 2005, 4:00 am Inst: Turner 1 This printout should have 13 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 3) 10 points A ski jumper travels down a slope and leaves the ski track moving in the horizontal direc tion with a speed of 26 . 4 m / s as in the figure. The landing incline below her falls off with a slope of = 31 . 7 . The acceleration of gravity is 9 . 8 m / s 2 . y x d v f Calculate the distance d she travels along the incline before landing. Correct answer: 103 . 251 m. Explanation: It is convenient to select the origin ( x = y = 0) at the beginning of the jump. Since v x = 26 . 4 m / s and v y = 0 m/s in this case, we have x = v x t y = v y t 1 2 g t 2 = 1 2 g t 2 . The distance d she travels along the incline before landing is related to her x and y coor dinates by x = d cos y = d sin . Substituting these expressions for x and y into the two equations above, we obtain d cos = v x t d sin = 1 2 g t 2 . Excluding t from these equations gives d = 2 v 2 sin g cos 2 = (2) (26 . 4 m / s) 2 sin 31 . 7 (9 . 8 m / s 2 ) cos 2 31 . 7 = 103 . 251 m . 002 (part 2 of 3) 10 points Determine how long the ski jumper is air borne. Correct answer: 3 . 32755 s. Explanation: Excluding d rather than t from the system above, we obtain t = 2 v tan g = (2) (26 . 4 m / s) tan(31 . 7 ) 9 . 8 m / s 2 = 3 . 32755 s . 003 (part 3 of 3) 10 points What is the relative angle with which the ski jumper hits the slope? Correct answer: 19 . 3075 . Explanation: The direction t of the velocity vector (rel ative to the positive x axis) at impact is t = arctan v y v x = arctan  32 . 61 m / s 26 . 4 m / s = 51 . 0075 , where v y = g t and v x = v . Therefore the relative angle of impact on the slope is =  t  = 51 . 0075  31 . 7 = 19 . 3075 . 004 (part 1 of 5) 10 points A stone at the end of a sling is whirled in a ver tical circle of radius 65 cm. At points A and B the string is inclined from the horizontal axis at an angle 49 and the stone has speed Overton, Mays Homework 7 Due: Feb 8 2005, 4:00 am Inst: Turner 2 3 . 9 m / s, as shown in the figure. The center of the circle is 1 . 13 m above level ground....
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This note was uploaded on 03/22/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Work

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