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: homework 06 FIERRO, JEFFREY Due: Feb 4 2008, 11:00 pm 1 Question 1, chap 4, sect 4. part 1 of 1 10 points A dart gun is fired while being held hori zontally at a height of 0 . 664 m above ground level, and at rest relative to the ground. The dart from the gun travels a horizontal dis tance of 2 . 48 m. A child holds the same gun in a horizontal position while sliding down a 53 . 6 incline at a constant speed of 2 . 25 m / s. The acceleration of gravity is 9 . 8 m / s 2 . 53 . 6 2 . 2 5 m / s x x 1 . 02m What horizontal distance x will the dart travel if the child fires the gun forward when it is 1 . 02 m above the ground? Correct answer: 2 . 48184 m (tolerance 1 %). Explanation: Let : x = 2 . 48 m , y = 0 . 664 m , v = 2 . 25 m / s = 53 . 6 , and y 1 = 1 . 02 m First, find the initial velocity of the dart when shot horizontally, at rest, 0 . 664 m above the ground. From y = v y t + 1 2 g t 2 , with v y =0, and down defined as positive, we have t = radicalbigg 2 y g = radicalBigg 2 (0 . 664 m) (9 . 8 m / s 2 ) = 0 . 368117 s , and x = v x t , thus v x = x t = (2 . 48 m) (0 . 368117 s) = 6 . 73699 m / s . Since we now know the initial horizontal velocity, we can find how far the dart will go if it is shot horizontally, y 1 = 1 . 02 m above the ground while sliding down the board at v = 2 . 25 m / s . Note: Now there is a y component to the initial velocity. v y = v sin = (2 . 25 m / s) sin 53 . 6 = 1 . 81101 m / s . From the following kinematic expression (re member down is positive), y = v y t + 1 2 a t 2 we have a quadratic equation in t 1 2 g t 2 + v y t y 1 = 0 , so t = v y radicalBigg v 2 y 4 parenleftbigg 1 2 g parenrightbigg ( y 1 ) g = 1 (9 . 8 m / s 2 ) braceleftbigg (1 . 81101 m / s) bracketleftBig (1 . 81101 m / s) 2 2 (9 . 8 m / s 2 ) ( . 664 m) bracketrightBig 1 / 2 bracerightbigg = 0 . 307456 s , the negative value must be rejected. The total horizontal velocity is v x = v x + v cos = (6 . 73699 m / s) + (2 . 25 m / s) cos53 . 6 = 8 . 07218 m / s . homework 06 FIERRO, JEFFREY Due: Feb 4 2008, 11:00 pm 2 and the horizontal distance traveled is x = v x t = (8 . 07218 m / s) (0 . 307456 s) = 2 . 48184 m . Question 2, chap 4, sect 4. part 1 of 1 10 points Salmon often jump waterfalls to reach their breeding grounds. The acceleration of gravity is 9 . 81 m / s 2 . Starting 1.54 m from a waterfall 0.387 m in height, at what minimum speed must a salmon jumping at an angle of 37 . 6 leave the water to continue upstream? Correct answer: 4 . 81608 m / s (tolerance 1 %). Explanation: Basic Concepts: Horizontally, x = v i (cos ) t Vertically, y = v i (sin ) t 1 2 g ( t ) 2 Given: x = 1 . 54 m y = 0 . 387 m = 37 . 6 g = 9 . 81 m / s 2 Solution: From the horizontal motion, t = x v i (cos ) so that y = v i (sin ) bracketleftbigg x v i (cos ) bracketrightbigg 1 2 g bracketleftbigg x v i (cos ) bracketrightbigg...
View
Full
Document
This homework help was uploaded on 04/03/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Work

Click to edit the document details