This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: quach (nyq56) Homework 2 sai (58015) 1 This printout should have 27 questions. Multiplechoice 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...
View
Full
Document
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.
 Spring '09
 NASAI
 Work

Click to edit the document details