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# HW6 - Pickering Tracey Homework 6 Due Feb 3 2006 noon Inst...

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Pickering, Tracey – Homework 6 – Due: Feb 3 2006, noon – Inst: Drummond 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points A jet airliner moving initially at 636 mph to the east where there is no wind moves into a region where the wind is blowing at 853 mph in a direction 15 north of east. What is the new speed of the aircraft with respect to the ground? Correct answer: 1476 . 53 mph. Explanation: V jet V wind V φ θ The new velocity of the aircraft is the sum of the initial velocity and the velocity of the wind: ~v f = ~v i + ~v wind = v x ˆ ı + v y ˆ where v x = v i + v wind cos φ v y = v wind sin φ with φ being the angle between v wind and v jet . So the speed is | ~v | = q v 2 x + v 2 y . 002 (part 2 of 2) 10 points What is the direction of the aircraft (use coun- terclockwise from due East as the positive an- gular direction, between the limits of - 180 and +180 )? Correct answer: 8 . 59917 . Explanation: The direction is θ = arctan v y v x . 003 (part 1 of 1) 10 points A dart gun is fired while being held hori- zontally at a height of 0 . 865 m above ground level, and at rest relative to the ground. The dart from the gun travels a horizontal dis- tance of 6 . 03 m. A child holds the same gun in a horizontal position while sliding down a 30 . 4 incline at a constant speed of 2 . 89 m / s. The acceleration of gravity is 9 . 8 m / s 2 . 30 . 4 2 . 89 m / s x x 1 . 26 m What horizontal distance x will the dart travel if the child fires the gun forward when it is 1 . 26 m above the ground? Correct answer: 6 . 39023 m. Explanation: Let : x 0 = 6 . 03 m , y 0 = 0 . 865 m , v 0 = 2 . 89 m / s θ = 30 . 4 , and y 1 = 1 . 26 m First, find the initial velocity of the dart when shot horizontally, at rest, 0 . 865 m above the ground. From y = v y 0 t + 1 2 g t 2 , with v y 0 =0, and down defined as positive, we have t = r 2 y 0 g = s 2 (0 . 865 m) (9 . 8 m / s 2 ) = 0 . 420155 s , and x = v x 0 t , thus v x 0 = x 0 t = (6 . 03 m) (0 . 420155 s) = 14 . 3518 m / s .

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Pickering, Tracey – Homework 6 – Due: Feb 3 2006, noon – Inst: Drummond 2 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 . 26 m above the ground while sliding down the board at v 0 = 2 . 89 m / s .
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HW6 - Pickering Tracey Homework 6 Due Feb 3 2006 noon Inst...

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