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Unformatted text preview: Version 098/ABCAC – midterm 01 – Turner – (58185) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 0 . 51 kg rock is projected from the edge of the top of a building with an initial velocity of 7 . 26 m / s at an angle 40 ◦ above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 10 . 1 m from the base of the building. Building 10 . 1 m 4 ◦ 7 . 2 6 m / s h How tall is the building? Assume the ground is level and that the side of the build ing is vertical. The acceleration of gravity is 9 . 8 m / s 2 . 1. 10.8971 2. 7.6857 3. 13.4055 4. 10.7813 5. 10.6073 6. 8.19745 7. 6.09122 8. 6.66079 9. 11.1946 10. 13.7102 Correct answer: 7 . 6857 m. Explanation: Let : θ = 40 ◦ , v = 7 . 26 m / s , Δ x = 10 . 1 m , and m = 0 . 51 kg . The flying time can be determined by x = v x t t = x v x = x v cos θ . From the point where the rock was projected (set to be the origin O ), the ycoordinate of the point where the rock struck the ground is y = v y t − 1 2 g t 2 = v sin θ parenleftbigg x v cos θ parenrightbigg − 1 2 g parenleftbigg x v cos θ parenrightbigg 2 = x tan θ − g x 2 2 ( v cos θ ) 2 , so h =  y  = g x 2 2 ( v cos θ ) 2 − x tan θ = (9 . 8 m / s 2 ) (10 . 1 m) 2 2 [(7 . 26 m / s) cos40 ◦ ] 2 − (10 . 1 m) tan40 ◦ = 7 . 6857 m . 002 10.0 points Consider the setup of a gun aimed at a target (such as a monkey) as shown in the figure. v θ d h x y O A B P 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 = 108 m / s, let the angle between the vector vectorv and the horizontal ( x ) direction be θ = 59 . 5 ◦ Version 098/ABCAC – midterm 01 – Turner – (58185) 2 and let AB = 81 . 8 m. The distance d = OB is the xcoordinate of the target. The acceleration of gravity is 9 . 8 m / s 2 . The height BP where the collision takes place is 1. 76.0792 2. 78.0137 3. 82.9508 4. 86.41 5. 81.9062 6. 91.499 7. 90.5456 8. 80.1792 9. 88.0344 10. 88.9324 Correct answer: 78 . 0137 m. 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 . The distance d is d = OB = AB tan θ = 81 . 8 m tan59 . 5 ◦ = 48 . 1838 m . t = d v cos θ = (48 . 1838 m) (108 m / s) cos(59 . 5 ◦ ) = 0 . 879041 s ....
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This note was uploaded on 03/26/2010 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner

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