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# hw2sol - homework 02 ARMSTRONG DOMINIC Due 3:00 am...

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homework 02 – ARMSTRONG, DOMINIC – Due: Sep 18 2007, 3:00 am 1 Question 1, chap 2, sect 6. part 1 of 3 10 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 . 321 s (see the figure). The points in the figure represent the sequential order, and are not drawn to scale. The distance AB = 0 . 87 m. O C B A 0 . 87 m x y Find the average speed as the ball passes the window. Correct answer: 2 . 71028 m / s (tolerance ± 1 %). Explanation: Basic Concepts: vectora = Δ vectorv Δ t For constant acceleration ¯ v = Δ s Δ t = v i + v f 2 v = v 0 + a t . For the acceleration of gravity ( a = g ) v = v 0 g t . Solution: The average velocity is given by ¯ v = Δ s Δ t = | AB | Δ t = 0 . 87 m 0 . 321 s = 2 . 71028 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 = 0 . 87 m + 1 2 (9 . 8 m / s 2 ) (0 . 321 s) 2 (0 . 321 s) = 4 . 28318 m / s v B = v A g t = 4 . 28318 m / s (9 . 8 m / s 2 ) (0 . 321 s) = 1 . 13738 m / s v = v A + v B 2 = 4 . 28318 m / s + 1 . 13738 m / s 2 = 2 . 71028 m / s . Question 2, chap 2, sect 6. part 2 of 3 10 points What is the magnitude of the decrease of the velocity from A to B? Correct answer: 3 . 1458 m / s (tolerance ± 1 %). 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 . 321 s) = 3 . 1458 m / s . Alternative Part 2: Δ v = v A v B = 4 . 28318 m / s 1 . 13738 m / s = 3 . 1458 m / s .

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homework 02 – ARMSTRONG, DOMINIC – Due: Sep 18 2007, 3:00 am 2 Question 3, chap 2, sect 6. part 3 of 3 10 points If the ball continues its path upward with- out obstruction, find the travel time between B and C, where point C is at the ball’s maxi- mum height. Correct answer: 0 . 116059 s (tolerance ± 1 %). 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 = 2 . 71028 m / s 3 . 1458 m / s 2 = 1 . 13738 m / s .
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hw2sol - homework 02 ARMSTRONG DOMINIC Due 3:00 am...

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