131. We make use of Eq. 4-24 and Eq. 4-25. (a) With x= 180 m, θo= 30º, and vo= 43 m/s, we obtain 22(9.8 m/s )(180 m)tan(30 )(180 m)11 m2(43 m/s) (cos30 )y=°−=−°or | | 11 my=. This implies the rise is roughly eleven meters above the fairway. (b) The horizontal component (in the absence of air friction) is unchanged, but the
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