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Unformatted text preview: h ( t ) =8 t + 8 = 0 , or at t = 1 sec. (b) The maximum height of the ball is the height h ( t ) at t = 1 sec, or h (1) =4 · 1 2 + 8 · 1 + 32 = 36 m . (c) The ball hits the ground at the ﬁrst value of t for which h ( t ) = 0, i.e., when h ( t ) =4 t 2 + 8 t + 32 =4( t 22 t8) =4( t4)( t + 2) = 0 . But t ≥ 0, so it must be that the ball hits the ground at t = 4 sec....
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This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.
 Fall '10
 KIHYUNHYUN
 Calculus

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