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# oldhw5 - oldhomework 05 Turner(58185 This print-out should...

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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 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 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 A v B = gt AB , v A + v B 2 = h t AB 2. v A v B = gt AB , v A + v B = h t AB 3. v B v A = g t AB , v A + v B 2 = h t AB cor- rect 4. v B v A = gt AB , v A + v B = 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 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 = 7 . 02 m / s, and c 2 = 3 . 57 m / s 2 . Find the ball’s initial speed v 0 at t 0 = 0 s. Correct answer: 7 . 02 m / s. Explanation: Basic Concepts: v = dx dt a = dv dt = d 2 x dt 2 Solution: The velocity is simply the derivative of y with respect to t : v = dy dt = 7 . 02 m / s 2(3 . 57 m / s 2 ) t, which at t = 0 is v 0 = 7 . 02 m / s .

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oldhw5 - oldhomework 05 Turner(58185 This print-out should...

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