HW03-solutions

# HW03-solutions - markowitz(am45362 HW03 distler(56295 This...

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markowitz (am45362) – HW03 – distler – (56295) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of4)10.0points 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 = 4 v 0 g 2. t max = 2 v 0 g 3. t max = 2 v 0 g correct 4. t max = 3 v 0 g 5. t max = v 0 g 6. t max = 1 2 v 0 g 7. t max = 1 4 v 0 g 8. t max = 1 2 v 0 g 9. t max = 2 3 v 0 g 10. t max = 1 3 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(part2of4)10.0points The maximum height h max of the cannonball is given by 1. h max = 3 v 2 0 g 2. h max = 1 3 v 2 0 g 3. h max = 2 v 2 0 g 4. h max = 1 4 v 2 0 g correct 5. h max = 2 v 2 0 g 6. h max = 1 2 v 2 0 g 7. h max = v 2 0 g 8. h max = 1 2 v 2 0 g 9. h max = 2 3 v 2 0 g 10. h max = 4 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(part3of4)10.0points 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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markowitz (am45362) – HW03 – distler – (56295) 2 2. v h max = 2 v 0 3. v h max = 1 2 v 0 4. v h max = 2 v 0 5. v h max = 1 2 v 0 correct 6. v h max = 1 4 v 0 7. v h max = 1 3 v 0 8. v h max = 4 v 0 9. v h max = 2 3 v 0 10. v h max = 3 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(part4of4)10.0points 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. 72 < θ 74 2. 76 < θ 78 3. 60 < θ 62 4. 68 < θ 70 5. 62 < θ 64 6. 70 < θ 72 7. 78 < θ 80 8. 64 < θ 66 9. 74 < θ 76 correct 10. 66 < θ 68 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(part1of2)10.0points You are standing on the edge of a ravine that is 14 . 3 m wide. You notice a cave on the opposite wall whose ceiling is 9 . 6 m below your feet. The cave is 4 . 4 m deep, and has a vertical back wall. You decide to kick a rock across the ravine into the cave. What initial horizontal velocity must you give the rock so that the rock barely misses the overhang? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 10 . 2164 m / s. Explanation: Since all the vertical motion is directed downward, we can consider down to be posi- tive. Let : w = 14 . 3 m , y = 9 . 6 m , and g = 9 . 8 m / s 2 . For the vertical motion, v oy = 0, so y = 1 2 g t 2 t 1 = radicalbigg 2 y g = radicalBigg 2(9 . 6 m) 9 . 8 m / s 2 = 1 . 39971 s .
markowitz (am45362) – HW03 – distler – (56295) 3 For the horizontal motion, a x = 0 and the ve- locity remains constant throughout the mo- tion, so the horizontal distance is defined by x = v ox t

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