hw05 - alexander(jra2623 homework 05 Turner(92510 This...

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alexander (jra2623) – homework 05 – Turner – (92510) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A tennis ball is dropped from 1 . 74 m above the ground. It rebounds to a height of 0 . 963 m. With what velocity does it hit the ground? The acceleration of gravity is 9 . 8 m / s 2 . (Let down be negative.) Correct answer: - 5 . 83986 m / s. Explanation: Let : h = - 1 . 74 m , h 1 = 0 . 963 m , and g = - 9 . 8 m / s 2 . Its final velocity is given by v 1 = - radicalbig 2 g h = - radicalBig 2( - 9 . 8 m / s 2 )( - 1 . 74 m) = - 5 . 83986 m / s . The velocity is negative because of the down- ward direction. 002 (part 2 of 3) 10.0 points With what velocity does it leave the ground? Correct answer: 4 . 34451 m / s. Explanation: As the ball rebounds, its motion is equiva- lent to dropping it from a height of h 1 under the influence of gravity, and its speed will be v 2 = radicalbig 2 g h 1 = radicalBig 2(9 . 8 m / s 2 )(0 . 963 m) = 4 . 34451 m / s . The velocity is positive because of the upward direction. 003 (part 3 of 3) 10.0 points If the tennis ball were in contact with the ground for 0 . 0104 s, find the acceleration given to the tennis ball by the ground. Correct answer: 979 . 267 m / s 2 . Explanation: While in contact with the ground, v o = v 1 and v f = v 2 , so its acceleration while in contact with the ground is a = Δ v Δ t = v f - v o Δ t = v 2 - v 1 t = 4 . 34451 m / s - ( - 5 . 83986 m / s) 0 . 0104 s = 979 . 267 m / s 2 .
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