38_Physics ProblemsTechnical Physics

38_Physics ProblemsTechnical Physics - Chapter 2 P2.42 39 1...

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Chapter 2 39 P2.42 We have yg t v t y fi i = ++ 1 2 2 0 490 800 300 2 = −− + .. . ms m 2 ch af tt . Solving for t , t = ±+ 640 588 980 . . Using only the positive value for t , we find that t = 179 . s . P2.43 (a) yyv t a t fii =+ 1 2 2 : 400 150 490 150 2 . . =( ) () v i and v i = 10 0 . ms upw a rd. (b) vv a t =+= = 10 0 9 80 1 50 4 68 . . m s v f = 468 . m s downward P2.44 The bill starts from rest v i = 0 and falls with a downward acceleration of m s 2 (due to gravity). Thus, in 0.20 s it will fall a distance of yv t g t i = = = 1 2 0 4 90 0 20 0 20 2 2 ... s m 2 . This distance is about twice the distance between the center of the bill and its top edge 8 cm a f . Thus, David will be unsuccessful .
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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