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Unformatted text preview: Version 112 – 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 10.0 points A target lies flat on the ground 6 m from the side of a building that is 10 m tall, as shown below. The acceleration of gravity is 10 m / s 2 . Air resistance is negligible. A student rolls a 6 kg ball off the horizontal roof of the building in the direction of the target. 6 m 10m v 10m The horizontal speed v with which the ball must leave the roof if it is to strike the target is most nearly 1. v = 6 √ 3 m/s. 2. v = 3 √ 6 m/s. 3. v = √ 5 6 m/s. 4. v = 6 √ 5 m/s. 5. v = 2 √ 3 m/s. 6. v = √ 3 6 m/s. 7. v = 3 √ 2 m/s. correct 8. v = 6 √ 2 m/s. 9. v = 2 √ 6 m/s. 10. v = √ 2 6 m/s. Explanation: m = 6 kg , not required h = 10 m , x = 6 m , and g = 10 m / s 2 . Observe the motion in the vertical direction only and it is a purely 1-dimension movement with a constant acceleration. So the time need for the ball to hit the ground is t = radicalBigg 2 h g and the horizontal speed should be v = x t for the ball to hit the target. Therefore v = x radicalbigg g 2 h = (6 m) radicalBigg 10 m / s 2 2 (10 m) = 6 √ 2 m / s = 3 √ 2 m / s . 002 (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 Version 112 – TEST01 – TSOI – (58160) 2 1. t max = v cos α g 2. t max = v g 3. t max = v g 4. t max = v sin α g correct 5. t max = v cos α 2 g 6. t max = v sin α 2 g 7. t max = g v 8. t max = 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....
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This note was uploaded on 05/04/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
- Spring '08