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Unformatted text preview: sanchez (ds28677) – homework 08 – Turner – (58220) 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 17 . 8 ◦ 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 3 . 47 m / s 2 and travels 69 . 3 m to the edge of the cliff. The cliff is 32 . 6 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 . 4363 m. Explanation: Given : a = 3 . 47 m / s 2 , l = 69 . 3 m , h = 32 . 6 m , and θ = 17 . 8 ◦ . 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 (3 . 47 m / s 2 ) (69 . 3 m) = 21 . 9304 m / s . Now, consider the projectile phase of the car’s motion. The initial vertical velocity is v iy = v sin θ = (21 . 9304 m / s) sin(17 . 8 ◦ ) = 6 . 70402 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 = (21 . 9304 m / s) 2 sin 2 17 . 8 ◦ + 2 ( 9 . 8 m / s 2 ) (32 . 6 m) = 683 . 904 m 2 / s 2 , so that v y = radicalBig 683 . 904 m 2 / s 2 = 26 . 1516 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 = (21 . 9304 m / s) cos 17 . 8 ◦ × (26 . 1516 m / s) (6 . 70402 m / s) (9 . 8 m / s 2 ) = 41 . 4363 m . 002 (part 2 of 2) 10.0 points Find the length of time the car is in the air. Correct answer: 1 . 98444 s. Explanation: Using the vertical motion, v y = v iy + g t t = v y v iy g = (26 . 1516 m / s) (6 . 70402 m / s) (9 . 8 m / s 2 ) = 1 . 98444 s ....
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This note was uploaded on 04/07/2011 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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