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ch02-p060 - 60 We choose down as the y direction and set...

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to that. The corresponding distance is y y' = 0.50 h , where y denotes the location of the ground. In these terms, y is the same as h , so we have h y' = 0.50 h or 0.50 h = y' . (a) We find t' and t from Eq. 2-15 (with v 0 = 0): 2 2 1 2 2 1 2 . 2 y y gt t g y y gt t g ′= ¡ ′= = ¡ = Plugging in y = h and y' = 0.50 h , and dividing these two equations, we obtain t t h g h g = = 2 050 2 050 . / / . . b g Letting t' = t – 1.00 (SI units understood) and cross-multiplying, we find t t t = ¡ = 100 050 100 1 050 . . . . which yields t = 3.41 s. (b) Plugging this result into y gt = 1 2 2 we find h = 57 m. (c) In our approach, we did not use the quadratic formula, but we did “choose a root” when we assumed (in the last calculation in part (a)) that 050
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