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**Unformatted text preview: **40. Neglect of air resistance justifies setting a = -g = -9.8 m/s2 (where down is our -y direction) for the duration of the fall. This is constant acceleration motion, and we may use Table 2-1 (with y replacing x). (a) Using Eq. 2-16 and taking the negative root (since the final velocity is downward), we have v=-
2 v0 - 2gy = - 0 - 2(9.8)(-1700) = -183 in SI units. Its magnitude is therefore 183 m/s. (b) No, but it is hard to make a convincing case without more analysis. We estimate the mass of a raindrop to be about a gram or less, so that its mass and speed (from part (a)) would be less than that of a typical bullet, which is good news. But the fact that one is dealing with many raindrops leads us to suspect that this scenario poses an unhealthy situation. If we factor in air resistance, the final speed is smaller, of course, and we return to the relatively healthy situation with which we are familiar. ...

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