hw2_solution - quach (nyq56) Homework 2 sai (58015) 1 This...

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Unformatted text preview: quach (nyq56) Homework 2 sai (58015) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A ball is thrown upward. Its initial verti- cal speed is 8 . 9 m / s , acceleration of gravity is 9 . 8 m / s 2 , and maximum height h max are shown in the figure below. Neglect: Air resistance. b b b b b b b b b b b b b b b b b b b b 8 . 9m / s 9 . 8m / s 2 h max What is its maximum height, h max ? Correct answer: 4 . 04133 m. Explanation: Let : v = 8 . 9 m / s and g = 9 . 8 m / s 2 . Basic Concept: For constant accelera- tion, we have v 2 = v 2 + 2 a ( y y ) . (1) Solution: The velocity at the top is zero. Since we know velocities and acceleration, Eq. 1 containing v , a , y , and y (no time) is the easiest one to use. Choose the positive direction to be up; then a = g and 0 = v 2 + 2 ( g ) ( h max 0) or h max = v 2 2 g = (8 . 9 m / s) 2 2 (9 . 8 m / s 2 ) = 4 . 04133 m . 002 (part 1 of 2) 10.0 points A ball has an initial speed of 25 m / s. The acceleration of gravity is 9 . 8 m / s 2 . t h y 3 . 3 s bbb b b b b b b b b b b b b b b b b b b b b b 25 m / s What will be its position after 3 . 3 s if it is thrown a) down with an initial speed of 25 m / s? Correct answer: 135 . 861 m. Explanation: Basic Concepts: This problem involves initial velocities directed both downward and upward, so all downward motion must be neg- ative. Assuming the initial position is h o = 0, the position at any time t is given by h f = v o t 1 2 g t 2 . Let : t = 3 . 3 s , v o = 25 m / s , and g = 9 . 8 m / s 2 . Solution: h 1 = v t 1 2 g t 2 = (25 m / s) (3 . 3 s) 1 2 (9 . 8 m / s 2 ) (3 . 3 s) 2 = 135 . 861 m . 003 (part 2 of 2) 10.0 points quach (nyq56) Homework 2 sai (58015) 2 t h y 3 . 3 s b b b b b b b b b b b b b b b b b b b b b b b b 25 m / s b) up with an initial speed of 25 m / s? Correct answer: 29 . 139 m. Explanation: h 2 = v t 1 2 g t 2 = (25 m / s) (3 . 3 s) 1 2 (9 . 8 m / s 2 ) (3 . 3 s) 2 = 29 . 139 m . 004 10.0 points You and your friend throw balloons filled with water from the roof of a several story apart- ment house. You simply drop a balloon from rest. A second balloon is thrown downward by your friend 2 . 7 s later with an initial speed of 52 . 92 m / s. They hit the ground simultane- ously. The acceleration of gravity is 9 . 8 m / s 2 . You can neglect air resistance. How high is the apartment house? Correct answer: 80 . 3723 m. Explanation: For balloon 1: y 1 = 1 2 g t 2 1 For balloon 2: y 2 = v t 2 + g t 2 2 2 Balloon 2 is thrown t seconds after balloon 1 and they hit the ground at the same time, so: t 1 = t 2 + t Since the height of the house is the same for both of the balloons, y 1 = y 2 g t 2 1 2 = v ( t 1 t ) + g 2 ( t 1 t ) 2 ( v g t ) t 1 = v t g 2 t 2 t 1 = v t g 2 t 2 v g t = t 2 + v v g t t 2 For...
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This note was uploaded on 04/25/2009 for the course PHY 58015 taught by Professor Nasai during the Spring '09 term at University of Texas at Austin.

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hw2_solution - quach (nyq56) Homework 2 sai (58015) 1 This...

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