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Unformatted text preview: homework 03 – GAUTHIER, JOSEPH – Due: Sep 21 2007, 1:00 am 1 Question 1, chap 4, sect 3. part 1 of 2 10 points A particle travels horizontally between two parallel walls separated by 18 . 4 m. It moves toward the opposing wall at a constant rate of 9 . 9 m / s. Also, it has an acceleration in the direction parallel to the walls of 4 . 1 m / s 2 . It hits the opposite wall at the same height. 18 . 4 m 4 . 1m / s 2 9 . 9 m / s a) What will be its speed when it hits the opposing wall? Correct answer: 12 . 4931 m / s (tolerance ± 1 %). Explanation: Let : d = 18 . 4 m , v x = 9 . 9 m / s , a = 4 . 1 m / s 2 , Basic Concepts Kinematics equations v = v o + g t s = s o + v o t + 1 2 g t 2 d a 7 . 6202m / s 9 . 9 m / s 1 2 . 4 9 3 1 m / s 5 2 . 4 1 3 9 ◦ 7 . 0814 m The horizontal motion will carry the parti cle to the opposite wall, so d = v x t f and t f = d v x = (18 . 4 m) (9 . 9 m / s) = 1 . 85859 s . is the time for the particle to reach the oppo site wall. Horizontally, the particle reaches the maxi mum parallel distance when it hits the oppo site wall at the time of t = d v x , so the final parallel velocity v y is v y = a t = a d v x = (4 . 1 m / s 2 ) (18 . 4 m) (9 . 9 m / s) = 7 . 6202 m / s . The velocities act at right angles to each other, so the resultant velocity is v f = radicalBig v 2 x + v 2 y = radicalBig (9 . 9 m / s) 2 + (7 . 6202 m / s) 2 = 12 . 4931 m / s . Question 2, chap 4, sect 3. part 2 of 2 10 points homework 03 – GAUTHIER, JOSEPH – Due: Sep 21 2007, 1:00 am 2 b) At what angle with the wall will the particle strike? Correct answer: 52 . 4139 ◦ (tolerance ± 1 %). Explanation: When the particle strikes the wall, the ver tical component is the side adjacent and the horizontal component is the side opposite the angle, so tan θ = v x v y , so θ = arctan parenleftbigg v x v y parenrightbigg = arctan parenleftbigg 9 . 9 m / s 7 . 6202 m / s parenrightbigg = 52 . 4139 ◦ . Note: The distance traveled parallel to the walls is y = 1 2 a t 2 = 1 2 (4 . 1 m / s 2 ) (1 . 85859 s) 2 = 7 . 0814 m . Question 3, chap 4, sect 3. part 1 of 1 10 points A quarterback takes the ball from the line of scrimmage, runs backward for 11 . 0 yards, then runs sideways parallel to the line of scrimmage for 11 . 0 yards. At this point, he throws a 30 . 0 yard forward pass straight down the field. What is the magnitude of the football’s resultant displacement? Correct answer: 21 . 9545 yards (tolerance ± 1 %). Explanation: Basic Concepts: Δ x tot = Δ x Δ y tot = Δ y 1 + Δ y 2 The displacements are perpendicular, so d tot = radicalBig (Δ x tot ) 2 + (Δ y tot ) 2 Given: Δ y 1 = − 11 . 0 yards Δ x = ± 11 . 0 yards Δ y 2 = 30 . 0 yards Solution: Δ y tot = − 11 yards + 30 yards = 19 yards Δ x tot = ± 11 yards Thus d = radicalBig ( ± 11 yards) 2 + (19 yards) 2 = 21 . 9545 yards Question 4, chap 4, sect 4....
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This note was uploaded on 03/19/2008 for the course M 408M taught by Professor Gilbert during the Fall '07 term at University of Texas.
 Fall '07
 Gilbert

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