to that. The corresponding distance is y– y'= 0.50h, where ydenotes the location of the ground. In these terms, yis the same as h, so we have h–y'= 0.50hor 0.50h= y'. (a) We find t' andtfrom Eq. 2-15 (with v0= 0): 2212212.2yygttgyygttg′′=′¡′==¡=Plugging in y= hand y'= 0.50h, and dividing these two equations, we obtain tthghg′==2 0502050.//..bgLetting t'= t– 1.00 (SI units understood) and cross-multiplying, we find ttt−=¡=−1000501001050....which yields t= 3.41 s. (b) Plugging this result into ygt=122we 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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Quadratic equation, one second, coordinate origin, 57 M