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Unformatted text preview: hernandez (ejh742) – homework 08 – Turner – (58120) 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 29 . 5 ◦ below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline with a constant acceleration of 6 . 08 m / s 2 and travels 47 . 9 m to the edge of the cliff. The cliff is 43 . 1 m above the ocean. The acceleration of gravity is 9 . 8 m / s 2 . Find the car’s position relative to the base of the cliff when the car lands in the ocean. Correct answer: 41 . 8314 m. Explanation: Given : a = 6 . 08 m / s 2 , l = 47 . 9 m , h = 43 . 1 m , and θ = 29 . 5 ◦ . The speed of the car when it reaches the edge of the cliff is given by v 2 = v 2 i + 2 a l = 2 a l since v i = 0 m / s, so v = √ 2 a ℓ = radicalBig 2 (6 . 08 m / s 2 ) (47 . 9 m) = 24 . 1343 m / s . Now, consider the projectile phase of the car’s motion. The initial vertical velocity is v iy = v sin θ = (24 . 1343 m / s) sin(29 . 5 ◦ ) = 11 . 8843 m / s . and the vertical velocity with which the car strikes the water is v 2 y = v 2 iy + 2 g h = v 2 sin 2 θ + 2 g h = (24 . 1343 m / s) 2 sin 2 29 . 5 ◦ + 2 ( 9 . 8 m / s 2 ) (43 . 1 m) = 985 . 997 m 2 / s 2 , so that v y = radicalBig 985 . 997 m 2 / s 2 = 31 . 4006 m / s . To find the time of flight, v y = v iy + g Δ t Δ t = v y v iy g The initial horizontal velocity is v ix = v cos θ and the horizontal motion of the car during this time is x = v ix Δ t = ( v cos θ ) v y v iy g = (24 . 1343 m / s) cos 29 . 5 ◦ × (31 . 4006 m / s) (11 . 8843 m / s) (9 . 8 m / s 2 ) = 41 . 8314 m . 002 (part 2 of 2) 10.0 points Find the length of time the car is in the air. Correct answer: 1 . 99146 s. Explanation: Using the vertical motion, v y = v iy + g t t = v y v iy g = (31 . 4006 m / s) (11 . 8843 m / s) (9 . 8 m / s 2 ) = 1 . 99146 s ....
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This note was uploaded on 06/14/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Work

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