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PHY303K HW2 - Perez(nap563 HW02 TSOI(58160 This print-out...

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Perez (nap563) – HW02 – TSOI – (58160) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points Denote the initial speed of a cannon ball fired from a battleship as v 0 . When the initial projectile angle is 45 with respect to the horizontal, it gives a maximum range R . y x θ 45 R R/ 2 The time of flight t max of the cannonball for this maximum range R is given by 1. t max = 2 3 v 0 g 2. t max = 1 2 v 0 g 3. t max = 2 v 0 g correct 4. t max = 1 4 v 0 g 5. t max = 1 3 v 0 g 6. t max = 3 v 0 g 7. t max = 1 2 v 0 g 8. t max = v 0 g 9. t max = 2 v 0 g 10. t max = 4 v 0 g Explanation: The cannonball’s time of flight is t = 2 v 0 y g = 2 v 0 sin 45 g = 2 v 0 g . 002 (part 2 of 4) 10.0 points The maximum height h max of the cannonball is given by 1. h max = 1 2 v 2 0 g 2. h max = 3 v 2 0 g 3. h max = 2 3 v 2 0 g 4. h max = 4 v 2 0 g 5. h max = 1 3 v 2 0 g 6. h max = 2 v 2 0 g 7. h max = 1 4 v 2 0 g correct 8. h max = v 2 0 g 9. h max = 1 2 v 2 0 g 10. h max = 2 v 2 0 g Explanation: Use the equation v 2 y = v 2 y 0 - 2 g h . At the top of its trajectory v y = 0. Solving for h yields h = v 2 y 0 2 g = v 2 0 sin 2 45 2 g = 1 4 v 2 0 g . 003 (part 3 of 4) 10.0 points The speed v h max of the cannonball at its max- imum height is given by 1. v h max = v 0
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Perez (nap563) – HW02 – TSOI – (58160) 2 2. v h max = 2 v 0 3. v h max = 1 3 v 0 4. v h max = 4 v 0 5. v h max = 2 3 v 0 6. v h max = 1 2 v 0 correct 7. v h max = 2 v 0 8. v h max = 1 4 v 0 9. v h max = 3 v 0 10. v h max = 1 2 v 0 Explanation: At the top of the cannonball’s trajectory, v y = 0. Hence the speed is equal to v x . | v | = v x = v 0 cos 45 = 1 2 v 0 . 004 (part 4 of 4) 10.0 points At a new angle, θ , the new range is given by R = R 2 . The corresponding angle, θ , which is greater than 45 , is given by 1. 78 < θ 80 2. 62 < θ 64 3. 68 < θ 70 4. 66 < θ 68 5. 70 < θ 72 6. 64 < θ 66 7. 76 < θ 78 8. 74 < θ 76 correct 9. 72 < θ 74 10. 60 < θ 62 Explanation: The maximum range corresponds to when θ = 45 R = v 2 0 sin[2 (45 )] g = v 2 0 g . Thus for R = R 2 , we need sin[2 θ ] = 1 2 . There are two solutions for θ that satisfy the above θ = 15 or θ = 75 . 005 (part 1 of 2) 10.0 points A car travels 11 . 8 km due north and then 38 . 5 km in a direction φ = 66 . 8 west of north. β θ φ A B R x y N E S W Find the magnitude of the car’s resultant displacement. Correct answer: 44 . 4907 km. Explanation: Let : A = 11 . 8 km , B = 38 . 5 km , and φ = 66 . 8 . θ = 180 - φ = 180 - 66 . 8 = 113 . 2 ,
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Perez (nap563) – HW02 – TSOI – (58160) 3 applying the Law of Cosines, R 2 = A 2 + B 2 - 2 A B cos θ . Since - 2 AB cos θ = - 2 (11 . 8 km)(38 . 5 km) cos 113 . 2 = 357 . 936 km 2 , then R = radicalBig (11 . 8 km) 2 + (38 . 5 km) 2 + 357 . 936 km 2 = 44 . 4907 km . 006 (part 2 of 2) 10.0 points Calculate the direction of the car’s resul- tant displacement, measured counterclock- wise from the northerly direction.
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