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**Unformatted text preview: **48. We neglect air resistance, which justifies setting a = -g = -9.8 m/s2 (taking down as the -y direction) for the duration of the motion. We are allowed to use Table 2-1 (with y replacing x) because this is constant acceleration motion. The ground level is taken to correspond to the origin of the y axis. The total time of fall can be computed from Eq. 2-15 (using the quadratic formula). v0 + 1 y = v0 t - gt2 = t = 2
2 v0 - 2gy g with the positive root chosen. With y = 0, v0 = 0 and y0 = h = 60 m, we obtain t= 2gh = g 2h = 3.5 s . g Thus, "1.2 s earlier" means we are examining where the rock is at t = 2.3 s: 1 y - h = v0 (2.3) - g(2.3)2 = y = 34 m 2 where we again use the fact that h = 60 m and v0 = 0. ...

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