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Unformatted text preview: Patel (pnp223) – homework 07 – Turner – (92160) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points Consider the setup of a gun aimed at a target (such as a monkey) as shown in the figure below. The target is to be dropped from the point A at t = 0, the same moment as the gun is fired. The bullet hits the target at a point P . Let the initial speed of the bullet be v = 107 m / s, let the angle between the vector vector v and the horizontal ( x) direction be θ = 53 . 6 ◦ and let AB = 92 . 1 m. The distance d = OB is the xcoordinate of the target. Denote the time taken to hit the target by t . v θ ℓ h x y O A B P The acceleration of gravity is 9 . 8 m / s 2 . This time is given by 1. t = d v cot θ . 2. t = d v cos θ . correct 3. t = d v tan θ . 4. t = d v sin θ . 5. t = d v . Explanation: Basic Concepts: Constant acceleration: x x = v t + 1 2 at 2 (1) v = v + at (2) Solution: Think of this twodimensional motion as two onedimensional trajectories, one in the x and one in the ydirection. The horizontal ( x ) initial velocity is v x = v cos θ . The xmotion is unaccelerated (gravity only affects the ymotion) so equation (1) gives d = v x t + 0 . Thus t = d v x = d v cos θ . 002 (part 2 of 4) 10.0 points Assume: Up is the positive direction. With the time taken to hit the target given by t , the vertical ( y) component of the veloc ity of the bullet is 1. v y = v sin θ g t 2. v y = g t 3. v y = v cos θ g t 4. v y = v cos θ g t 5. v y = v sin θ + g t 6. v y = v sin θ g t correct 7. v y = v sin θ + g t 8. v y = v cos θ + g t 9. v y = v cos θ + g t 10. v y = g t Explanation: The vertical ( y ) initial velocity is given by v y = v sin θ We take positive direction upward, so...
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This note was uploaded on 07/11/2009 for the course PHY 92160 taught by Professor Turner during the Spring '09 term at University of Texas.
 Spring '09
 Turner
 Physics, Work

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