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Unformatted text preview: adamkiewicz (ja28724) Motion in two dimentions Morozova (11102) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A cannon fires a 0 . 42 kg shell with initial velocity v i = 8 . 5 m / s in the direction = 40 above the horizontal. x h 8 . 5 m / s 4 y y The shells trajectory curves downward be- cause of gravity, so at the time t = 0 . 563 s the shell is below the straight line by some vertical distance h . Find this distance h in the absence of air resistance. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1 . 55315 m. Explanation: In the absence of gravity, the shell would fly along the straight line at constant velocity: x = t v i cos , y = t v i sin . The gravity does not affect the x coordinate of the shell, but it does pull its y coordinate downware at a constant acceleration a y =- g , so x = t v i cos , y = t v i sin - g t 2 2 . Thus, x = x but y = y- 1 2 gt 2 ; in other words, the shell deviates from the straight-line path by the vertical distance h = y- y = g t 2 2 . Note: This result is completely indepen- dent of the initial velocity v i or angle of the shell. It is a simple function of the flight time t . h = g t 2 2 = (9 . 8 m / s 2 ) (0 . 563 s) 2 2 = 1 . 55315 m . 002 (part 1 of 2) 10.0 points An artillery shell is fired at an angle of 66 . 6 above the horizontal ground with an initial speed of 1960 m / s. The acceleration of gravity is 9 . 8 m / s 2 . Find the total time of flight of the shell, neglecting air resistance. Correct answer: 6 . 11838 min. Explanation: Using half the flight we have v y = v sin and 0 = v fy = v sin + a y t, so that t = v sin g ....
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- Spring '08