# Test 1 - Version 022 – TEST01 – TSOI –(58160 1 This...

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Unformatted text preview: Version 022 – TEST01 – TSOI – (58160) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A projectile is fired with an initial speed v at t = 0. The angle between the initial velocity v and the horizontal plane is α . x y v α y max R A B The time t max it takes for the projectile to reach its maximum height is 1. t max = v sin α 2 g 2. t max = v sin α g correct 3. t max = v 4. t max = v cos α 2 g 5. t max = v g 6. t max = v g 7. t max = v cos α g 8. t max = g v Explanation: Basic Concepts: For two dimensional projectile motion in a gravitational field the acceleration is due to gravity only and acts exclusively on the y component of velocity. Kinematic equations for constant acceleration are applicable. Velocity in the x-direction is a constant. Solution: The initial y-velocity is v y = v sin α . The projectile reaches its maximum height when v y = 0. Under this condition the basic equation v = v + at, reduces to v y = 0 = v y − g t, so t max = v y g = v sin α g . 002 (part 2 of 3) 10.0 points Given v = 15 . 1 m / s and α = 30 ◦ , what is the speed of the projectile when it reaches its maximum height y = y max ; ( i.e. , at point A in the figure)? 1. 14.4626 2. 16.3679 3. 18.2731 4. 18.7928 5. 17.2339 6. 20.1784 7. 19.9186 8. 15.6751 9. 15.3286 10. 13.077 Correct answer: 13 . 077 m / s. Explanation: At its maximum height v y = 0, thus the only component of velocity is in the x- direction. The speed is thus v = v x = v cos α = (15 . 1 m / s) cos30 ◦ = 13 . 077 m / s . 003 (part 3 of 3) 10.0 points Find the speed of the projectile, on its way down, at the height y = y max 2 ( i.e. at point B in the figure). 1. 19.9243 2. 14.9666 3. 14.1248 4. 16.2762 5. 19.0825 6. 20.2985 7. 22.4499 8. 14.7795 9. 20.6727 Version 022 – TEST01 – TSOI – (58160) 2 10. 21.3274 Correct answer: 14 . 1248 m / s. Explanation: At B , v B = radicalBig v 2 xB + v 2 yB . The horizontal component of velocity is constant, so v xB = v x = v cos α. (1) The trick is finding v yB . Applying v 2 = v 2 + 2 as for the y-component of motion from A to B gives v 2 yB = 0 + 2 g parenleftbigg h − h 2 parenrightbigg = g h. (2) We need to determine h . Applying v 2 = v 2 + 2 as for the y − component of motion from...
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## This note was uploaded on 11/21/2010 for the course PHY 58160 taught by Professor Tsio during the Fall '10 term at University of Texas.

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Test 1 - Version 022 – TEST01 – TSOI –(58160 1 This...

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