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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 y gt v t y f i i = + + 1 2 2 0 4 90 8 00 30 0 2 = + . . . m s m s m 2 c h a f t t . Solving for t , t = ± + 8 00 64 0 588 9 80 . . . . Using only the positive value for t , we find that t = 1 79 . s . P2.43 (a) y y v t at f i i = + 1 2 2 : 4 00 1 50 4 90 1 50 2 . . . . =( ) ( )( ) v i and v i = 10 0 . m s upward . (b) v v at f i = + = ( )( )= 10 0 9 80 1 50 4 68 . . . . m s v f = 4 68 . m s downward P2.44 The bill starts from rest v i = 0 and falls with a downward acceleration of 9 80 . m s 2 (due to gravity). Thus, in 0.20 s it will fall a distance of y v t gt i = = ( ) = 1 2 0 4 90 0 20 0 20 2 2 . . . m s s m 2 c h . 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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