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**Unformatted text preview: **42. We neglect air resistance, which justifies setting a = -g = -9.8 m/s2 (taking down as the -y direction) for the duration of the fall. This is constant acceleration motion, which justifies the use of Table 2-1 (with y replacing x). (a) Noting that y = y - y0 = -30 m, we apply Eq. 2-15 and the quadratic formula (Appendix E) to compute t: 2 v0 v0 - 2gy 1 y = v0 t - gt2 = t = 2 g which (with v0 = -12 m/s since it is downward) leads, upon choosing the positive root (so that t > 0), to the result: -12 + (-12)2 - 2(9.8)(-30) t= = 1.54 s . 9.8 (b) Enough information is now known that any of the equations in Table 2-1 can be used to obtain v; however, the one equation that does not use our result from part (a) is Eq. 2-16: v=
2 v0 - 2gy = 27.1 m/s where the positive root has been chosen in order to give speed (which is the magnitude of the velocity vector). ...

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