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Unformatted text preview: markowitz (am45362) – HW03 – distler – (56295) 1 This printout should have 22 questions. Multiplechoice 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 . 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 g 2. t max = 2 v g 3. t max = √ 2 v g correct 4. t max = √ 3 v g 5. t max = v g 6. t max = 1 √ 2 v g 7. t max = 1 4 v g 8. t max = 1 2 v g 9. t max = 2 3 v g 10. t max = 1 √ 3 v g Explanation: The cannonball’s time of flight is t = 2 v y g = 2 v sin 45 ◦ g = √ 2 v g . 002 (part 2 of 4) 10.0 points The maximum height h max of the cannonball is given by 1. h max = √ 3 v 2 g 2. h max = 1 √ 3 v 2 g 3. h max = √ 2 v 2 g 4. h max = 1 4 v 2 g correct 5. h max = 2 v 2 g 6. h max = 1 2 v 2 g 7. h max = v 2 g 8. h max = 1 √ 2 v 2 g 9. h max = 2 3 v 2 g 10. h max = 4 v 2 g Explanation: Use the equation v 2 y = v 2 y 2 g h. At the top of its trajectory v y = 0. Solving for h yields h = v 2 y 2 g = v 2 sin 2 45 ◦ 2 g = 1 4 v 2 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 markowitz (am45362) – HW03 – distler – (56295) 2 2. v h max = 2 v 3. v h max = 1 2 v 4. v h max = √ 2 v 5. v h max = 1 √ 2 v correct 6. v h max = 1 4 v 7. v h max = 1 √ 3 v 8. v h max = 4 v 9. v h max = 2 3 v 10. v h max = √ 3 v Explanation: At the top of the cannonball’s trajectory, v y = 0. Hence the speed is equal to v x .  v  = v x = v cos 45 ◦ = 1 √ 2 v . 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. 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 sin[2 (45 ◦ )] g = v 2 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 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....
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This note was uploaded on 09/19/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Fall '08 term at University of Texas.
 Fall '08
 Kaplunovsky

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