(b) And we use Eq. 4-22 to solve for the initial height y0:222000011(sin) 0( 40.3 m/s)(10.0 s)(9.80 m/s )(10.0 s)yy vt gtyθ−=−¡−−which yields y0= 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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