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175_Physics ProblemsTechnical Physics

# 175_Physics ProblemsTechnical Physics - Chapter 6 P6.49(a F...

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Chapter 6 177 P6.49 (a) At terminal speed, F mg Cv = − + = 2 0. Thus, C mg v = = = × 2 2 4 0 046 0 9 80 44 0 2 33 10 . . . . kg m s m s kg m 2 b g e j b g (b) We set up a spreadsheet to calculate the motion, try different initial speeds, and home in on 53 m s as that required for horizontal range of 155 m, thus: Time t (s) x (m) v x (m/s) a x m s 2 e j y (m) v y (m/s) a y m s 2 e j v v v x y = + 2 2 (m/s) tan F H G I K J 1 v v y x (deg) 0.000 0 0.000 0 45.687 0 –10.565 9 0.000 0 27.451 5 –13.614 6 53.300 0 31.000 0 0.002 7 0.121 1 45.659 0 –10.552 9 0.072 7 27.415 5 –13.604 6 53.257 4 30.982 2 2.501 6 90.194 6 28.937 5 –4.238 8 32.502 4 0.023 5 –9.800 0 28.937 5 0.046 6 2.504 3 90.271 3 28.926 3 –4.235 5 32.502 4 –0.002 4 –9.800 0 28.926 3 –0.004 8 2.506 9 90.348 0 28.915 0 –4.232 2 32.502 4 –0.028 4 –9.800 0 28.915 1 –0.056 3 3.423 8 115.229 8 25.492 6 –3.289 6 28.397 2 –8.890 5 –9.399 9 26.998 4 –19.226 2 3.426 5 115.297 4 25.483 9 –3.287 4 28.373 6 –8.915 4 –9.397 7 26.998 4 –19.282 2 3.429 1 115.364 9 25.475 1 –3.285 1 28.350 0 –8.940 3 –9.395 4 26.998 4 –19.338 2 5.151 6 154.996 8 20.843 8 –2.199 2 0.005 9 –23.308 7 –7.049 8 31.269 2 –48.195 4 5.154 3 155.052 0 20.838 0 –2.198 0 –0.055 9 –23.327 4 –7.045 4 31.279 2 –48.226 2 (c) Similarly, the initial speed is 42 m s . The motion proceeds thus: Time t (s) x (m) v x (m/s) a x m s 2 e j y (m) v y (m/s) a y m s 2 e j v v v x y = + 2 2 (m/s) tan F H G I K J 1 v v y x (deg) 0.000 0 0.000 0 28.746 2 –4.182 9
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