This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: tokarski (at23678) Motion in two dimentions Morozova (11104) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A cannon fires a 0 . 308 kg shell with initial velocity v i = 10 m / s in the direction = 59 above the horizontal. x h 1 m / s 5 9 y y The shells trajectory curves downward be cause of gravity, so at the time t = 0 . 347 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: 0 . 590004 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 straightline 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 . 347 s) 2 2 = . 590004 m . 002 (part 1 of 2) 10.0 points An artillery shell is fired at an angle of 34 . 2 above the horizontal ground with an initial speed of 1970 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: 3 . 76635 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 ....
View
Full
Document
This document was uploaded on 06/01/2009.
 Spring '08

Click to edit the document details