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Unformatted text preview: adamkiewicz (ja28724) – Motion in two dimentions – Morozova – (11102) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A cannon fires a 0 . 42 kg shell with initial velocity v i = 8 . 5 m / s in the direction θ = 40 ◦ above the horizontal. Δ x Δ h 8 . 5 m / s 4 ◦ Δ y y The shell’s trajectory curves downward be cause of gravity, so at the time t = 0 . 563 s the shell is below the straight line by some vertical distance Δ h . Find this distance Δ h in the absence of air resistance. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1 . 55315 m. Explanation: In the absence of gravity, the shell would fly along the straight line at constant velocity: ˆ x = t v i cos θ , ˆ y = t v i sin θ . The gravity does not affect the x coordinate of the shell, but it does pull its y coordinate downware at a constant acceleration a y = g , so x = t v i cos θ, y = t v i sin θ g t 2 2 . Thus, x = ˆ x but y = ˆ y 1 2 gt 2 ; in other words, the shell deviates from the straightline path by the vertical distance Δ h = ˆ y y = g t 2 2 . Note: This result is completely indepen dent of the initial velocity v i or angle θ of the shell. It is a simple function of the flight time t . Δ h = g t 2 2 = (9 . 8 m / s 2 ) (0 . 563 s) 2 2 = 1 . 55315 m . 002 (part 1 of 2) 10.0 points An artillery shell is fired at an angle of 66 . 6 ◦ above the horizontal ground with an initial speed of 1960 m / s. The acceleration of gravity is 9 . 8 m / s 2 . Find the total time of flight of the shell, neglecting air resistance. Correct answer: 6 . 11838 min. Explanation: Using half the flight we have v y = v sin θ and 0 = v fy = v sin θ + a y t, so that t = v sin θ g ....
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This note was uploaded on 04/14/2009 for the course PHYS 111 taught by Professor Moro during the Spring '08 term at NJIT.
 Spring '08
 moro
 Physics

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