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Unformatted text preview: tokarski (at23678) – Motion in two dimentions – Morozova – (11104) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A cannon fires a 0 . 308 kg shell with initial velocity v i = 10 m / s in the direction θ = 59 ◦ above the horizontal. Δ x Δ h 1 m / s 5 9 ◦ Δ y y The shell’s trajectory curves downward be- cause of gravity, so at the time t = 0 . 347 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: 0 . 590004 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 straight-line 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 . 347 s) 2 2 = . 590004 m . 002 (part 1 of 2) 10.0 points An artillery shell is fired at an angle of 34 . 2 ◦ above the horizontal ground with an initial speed of 1970 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: 3 . 76635 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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- Spring '08
- Acceleration, 0 2g, 0 2g, 0 3g