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Unformatted text preview: hernandez (ejh742) oldhomework 05 Turner (58120) 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 ball is dropped from rest at point O . It passes a window with height 3 . 8 m in time interval t AB = 0 . 02 s. Assume: Down is the positive y direction. Use g = 9 . 8 m / s 2 . O A B 3 . 8 m b b b b b b b b b b b x y bardbl vectorv bardbl = v is the speed of the ball. Identify the correct pair of equations, which enable us to solve for speed v B . 1. v B v A = g t AB , v A + v B = h t AB 2. v B v A = g t AB , v A + v B 2 = h t AB cor rect 3. v A v B = g t AB , v A + v B = h t AB 4. v A v B = g t AB , v A + v B 2 = h t AB Explanation: Let : h = 3 . 8 m , t AB = 0 . 02 s , and g = 9 . 8 m / s 2 . Basic Concept The change in speed over time is acceleration by definition g = v B v A t AB . Note: The change in speed is the difference between the final and the initial speed over a time interval. O A B h t AB b b b b b b b b b b b b t y Solution v B is the final speed over the time needed to pass the window, so it appears first in the expression. The average speed for the ball can be expressed in terms of a dis placement over time or as an average of the final and initial speed (provided the accelera tion is constant); thus v Avg = v A + v B 2 = h t AB . 002 (part 1 of 3) 10.0 points The position of a softball tossed vertically upward is described by the equation y = c 1 t c 2 t 2 , where y is in meters, t in seconds, c 1 = 10 . 3 m / s, and c 2 = 2 . 28 m / s 2 ....
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This note was uploaded on 06/14/2011 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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