hw05 - alexander(jra2623 – homework 05 – Turner...

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Unformatted text preview: 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....
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This note was uploaded on 10/20/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Summer '08 term at University of Texas.

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hw05 - alexander(jra2623 – homework 05 – Turner...

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