ch04-p027 - 27. We adopt the positive direction choices...

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(b) And we use Eq. 4-22 to solve for the initial height y 0 : 2 22 00 0 0 11 ( sin ) 0 ( 40.3 m/s)(10.0 s) (9.80 m/s )(10.0 s) yy v t g t y θ −= ¡ which yields y 0 = 897 m. 27. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is at ground level directly below the release point. We write 0 = –30.0° since the angle shown in the figure is measured
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