PHY 303K - Homework 6 - homework 06 FIERRO, JEFFREY Due:...

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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...
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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.

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PHY 303K - Homework 6 - homework 06 FIERRO, JEFFREY Due:...

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