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Unformatted text preview: Niu (qn269) – HW02 – Tsoi – (58020) 1 This printout should have 22 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points Given: The acceleration of gravity on Earth is 9 . 8 m / s 2 . Consider a ball thrown up from the ground (the point O). It passes a window (the seg ment AB) in the time interval 0 . 262 s (see the figure). The points in the figure represent the sequential order, and are not drawn to scale. The distance AB = 1 . 17 m. O C B A 1 . 17 m b b b b b b b b b b b x y Find the average speed as the ball passes the window. Correct answer: 4 . 46565 m / s. Explanation: Basic Concepts: vectora = Δ vectorv Δ t For constant acceleration ¯ v = Δ s Δ t = v i + v f 2 v = v + a t . For the acceleration of gravity ( a = − g ) v = v − g t . Solution: The average velocity is given by ¯ v = Δ s Δ t =  AB  Δ t = 1 . 17 m . 262 s = 4 . 46565 m / s . Alternative Part 1: d = y B − y A = v A t − 1 2 g t 2 v A = d + 1 2 g t 2 t = 1 . 17 m + 1 2 (9 . 8 m / s 2 ) (0 . 262 s) 2 (0 . 262 s) = 5 . 74945 m / s v B = v A − g t = 5 . 74945 m / s − (9 . 8 m / s 2 ) (0 . 262 s) = 3 . 18185 m / s v = v A + v B 2 = 5 . 74945 m / s + 3 . 18185 m / s 2 = 4 . 46565 m / s . 002 (part 2 of 3) 10.0 points What is the magnitude of the decrease of the velocity from A to B? Correct answer: 2 . 5676 m / s. Explanation: For a constant acceleration vectora = Δ vectorv Δ t , = ⇒ Δ vectorv  =  vectora Δ t  = g Δ t . Thus the decrease in velocity is Δ v = g Δ t = (9 . 8 m / s 2 )(0 . 262 s) = 2 . 5676 m / s . Alternative Part 2: Δ v = v A − v B = 5 . 74945 m / s − 3 . 18185 m / s = 2 . 5676 m / s . Niu (qn269) – HW02 – Tsoi – (58020) 2 003 (part 3 of 3) 10.0 points If the ball continues its path upward without obstruction, find the travel time between B and C, where point C is at the ball’s maximum height. Correct answer: 0 . 324678 s. Explanation: The velocity change is v A − v B = Δ v = ⇒ v A = v B + Δ v and the average velocity is ¯ v = v A + v B 2 = 2 v B + Δ v 2 . Thus v B = ¯ v − Δ v 2 = 4 . 46565 m / s − 2 . 5676 m / s 2 = 3 . 18185 m / s . The ball reaches its maximum height when its velocity is zero. Based on v = v + a t v C = 0 = v B − g t t = v B g = 3 . 18185 m / s 9 . 8 m / s 2 = . 324678 s . 004 (part 1 of 2) 10.0 points A ball is thrown upward. Its initial verti cal speed v , acceleration of gravity g , and maximum height h max are shown in the figure below. Neglect: Air resistance. The acceleration of gravity is 9 . 8 m / s 2 . b b b b b b b b b b b b b b b b b b b b v 9 . 8m / s 2 h max What is its maximum height, h max (in terms of the initial speed v )?...
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This note was uploaded on 09/01/2009 for the course PHY M taught by Professor Staff during the Spring '09 term at University of Texas.
 Spring '09
 Staff
 Physics, Acceleration, Gravity

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