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AmTran_Bus

  • one year ago

I really need help.

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  1. AmTran_Bus
    • one year ago
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    From a hilltop to the west of Willy, Oompa Loompas launch gum balls out of slingshots toward his canoe. These gum balls travel in a linear path, dropping 3.2 meters as they travel eastward 15.3 meters to strike the canoe. 1. What's the distance from the Oompa Loompas to the canoe? 2. What is the angle, measured from the horizon, that the Oompa Loompas are firing from? 3. Is it possible for a sling sot propelled gumball to travel in a linear path? If not, what is the flaw in this logic?

  2. AmTran_Bus
    • one year ago
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    If I have it drawn right, part a is the pyth. theorem.

  3. AmTran_Bus
    • one year ago
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    |dw:1360440654830:dw|

  4. AmTran_Bus
    • one year ago
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    Maybe law of sines for part two?

  5. klimenkov
    • one year ago
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    I am not very good in English, but if your pic is correct, you found the distance right.

  6. AmTran_Bus
    • one year ago
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    Thanks so much! I think I drew it right. @Jonask what do you think?

  7. Jonask
    • one year ago
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    yes for 2|dw:1360448299629:dw|

  8. AmTran_Bus
    • one year ago
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    Awesome. Could I take the arctan?

  9. AmTran_Bus
    • one year ago
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    Whoops, no. I ment the law of sines.

  10. klimenkov
    • one year ago
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    \(\theta\) on the @Jonask 's pic is not measured from the horizon.

  11. Spacelimbus
    • one year ago
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    No matter which angle you search, the law of sines is a ratio, it is mainly used to to find specific sides, but in such a right-angled triangle, as it seems (only judging by the picture - I haven't fully read the problem) you wouldn't want to use that law, more likely the regular trigonometric functions and solve for the angle.

  12. AmTran_Bus
    • one year ago
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    Where is the horizon?

  13. AmTran_Bus
    • one year ago
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    I wonder if it is a right triangle, from part 4.

  14. klimenkov
    • one year ago
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    |dw:1360441416725:dw|

  15. AmTran_Bus
    • one year ago
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    Oh boy.

  16. Jonask
    • one year ago
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    yes its like an angle of elevation @klimenkov |dw:1360448668700:dw|

  17. AmTran_Bus
    • one year ago
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    yippee. So where do I go from here?

  18. AmTran_Bus
    • one year ago
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    1st of all, is the pic 100% correct?

  19. AmTran_Bus
    • one year ago
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    I really think the drawing is right. I really think the distance is right I do not know about part two, how to get the angle And I am unsure about the logic part 4.

  20. AmTran_Bus
    • one year ago
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    *part 3

  21. AmTran_Bus
    • one year ago
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    Anybody?

  22. Jonask
    • one year ago
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    tan theta=3.2/15.3

  23. Jonask
    • one year ago
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    not theta but alpha\[\tan \alpha=3.2/15.3\]

  24. AmTran_Bus
    • one year ago
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    That is 0.00365.

  25. Jonask
    • one year ago
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    now to get alpha use arctan(0.00365)

  26. AmTran_Bus
    • one year ago
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    That is 0.209129

  27. AmTran_Bus
    • one year ago
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    Thanks @Jonask

  28. AmTran_Bus
    • one year ago
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    Now for the next part///

  29. Jonask
    • one year ago
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    first of all gravity wont allow the ball just to down at the same angle without pulling it towards the earth 2nd air resistance especially because the horizontal distance is too big compared to the vertical

  30. AmTran_Bus
    • one year ago
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    I was thinking the same, had it wrote down. Wanted to verify. Thanks SOO much.

  31. AmTran_Bus
    • one year ago
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    So what needed to be done to find the angle from the horizon?

  32. Jonask
    • one year ago
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    its that arctan 0.00365=0.209 degrees this also accounts for the balls imppsossible perpetual linear motion

  33. AmTran_Bus
    • one year ago
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    Thanks and God Bless. Keep on living for Jesus.

  34. Jonask
    • one year ago
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    God Bless you too @AmTran_Bus

  35. Jonask
    • one year ago
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    @klimenkov

  36. klimenkov
    • one year ago
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    Looks correct.

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