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**Unformatted text preview: **46. 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 y = 0. (a) With y0 = h and v0 replaced with -v0 , Eq. 2-16 leads to v= (-v0 ) - 2g (y - y0 ) = 2 2 v0 + 2gh . The positive root is taken because the problem asks for the speed (the magnitude of the velocity). (b) We use the quadratic formula to solve Eq. 2-15 for t, with v0 replaced with -v0 , -v0 + 1 y = -v0 t - gt2 = t = 2 (-v0 ) - 2gy g 2 where the positive root is chosen to yield t > 0. With y = 0 and y0 = h, this becomes t= 2 v0 + 2gh - v0 . g (c) If it were thrown upward with that speed from height h then (in the absence of air friction) it would return to height h with that same downward speed and would therefore yield the same final speed (before hitting the ground) as in part (a). An important perspective related to this is treated later in the book (in the context of energy conservation) . (d) Having to travel up before it starts its descent certainly requires more time than in part (b). The calculation is quite similar, however, except for now having +v0 in the equation where we had put in -v0 in part (b). The details follow: v0 + 1 y = v0 t - gt2 = t = 2 2 v0 - 2gy g with the positive root again chosen to yield t > 0. With y = 0 and y0 = h, we obtain t= 2 v0 + 2gh + v0 . g ...

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