# sol3 - Homework Solution 3 1 question E5 page 56 v = 12 m s...

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Unformatted text preview: Homework Solution 3 1. question E5 page 56 v =- 12 m / s g =- 10 m / s 2 t = 1 . 0 s v = v + g t =- 12 m / s + (- 10 m / s 2 ) (1 . 0 s) =- 22 m / s . Here, “- ” means downward. 2. question E10 page 56 a. v = 18 m / s g =- 3 m / s 2 t = 4 s v = v + g t = 18 m / s + (- 3 m / s 2 ) (4 s) = 6 m / s . b. When the ball reaches the high point in its flight, it has a speed of 0. Thus v = v + g t 0 = 18 m / s + (- 3 m / s 2 ) t t = 6 s . 3. problem SP1 page 56 a. v = 0 when the ball reaches the high point. b. We know v = 16 m / s, g =- 10 m / s 2 and v = 0, so v = v + g t 0 = 16 m / s + (- 10 m / s 2 ) t t = 1 . 6 s . 1 c. d = v t + 1 2 g t 2 = (16 m / s) (1 . 6 s) + 1 2 (- 10 m / s 2 ) (1 . 6 s) 2 = 12 . 8 m . d. When t = 2 s, d = v t + 1 2 g t 2 = (16 m / s) (2 s) + 1 2 (- 10 m / s 2 ) (2 s) 2 = 12 m . e. When t = 2 s, v = v + g t = 16 m / s + (- 10 m / s) (2 s) =- 4 m / s . So the ball is moving down ....
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## This note was uploaded on 09/23/2009 for the course PHY 59335 taught by Professor Fischeller during the Fall '09 term at University of Texas.

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sol3 - Homework Solution 3 1 question E5 page 56 v = 12 m s...

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