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AmTran_Bus

  • 2 years ago

I really need help.

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

  3. AmTran_Bus
    • 2 years ago
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    |dw:1360440654830:dw|

  4. AmTran_Bus
    • 2 years ago
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    Maybe law of sines for part two?

  5. klimenkov
    • 2 years 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
    • 2 years ago
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    Thanks so much! I think I drew it right. @Jonask what do you think?

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

  8. AmTran_Bus
    • 2 years ago
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    Awesome. Could I take the arctan?

  9. AmTran_Bus
    • 2 years ago
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    Whoops, no. I ment the law of sines.

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

  11. Spacelimbus
    • 2 years 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
    • 2 years ago
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    Where is the horizon?

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

  14. klimenkov
    • 2 years ago
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    |dw:1360441416725:dw|

  15. AmTran_Bus
    • 2 years ago
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    Oh boy.

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

  17. AmTran_Bus
    • 2 years ago
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    yippee. So where do I go from here?

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

  19. AmTran_Bus
    • 2 years 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
    • 2 years ago
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    *part 3

  21. AmTran_Bus
    • 2 years ago
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    Anybody?

  22. Jonask
    • 2 years ago
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    tan theta=3.2/15.3

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

  24. AmTran_Bus
    • 2 years ago
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    That is 0.00365.

  25. Jonask
    • 2 years ago
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    now to get alpha use arctan(0.00365)

  26. AmTran_Bus
    • 2 years ago
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    That is 0.209129

  27. AmTran_Bus
    • 2 years ago
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    Thanks @Jonask

  28. AmTran_Bus
    • 2 years ago
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    Now for the next part///

  29. Jonask
    • 2 years 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
    • 2 years ago
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    I was thinking the same, had it wrote down. Wanted to verify. Thanks SOO much.

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

  32. Jonask
    • 2 years 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
    • 2 years ago
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    Thanks and God Bless. Keep on living for Jesus.

  34. Jonask
    • 2 years ago
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    God Bless you too @AmTran_Bus

  35. Jonask
    • 2 years ago
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    @klimenkov

  36. klimenkov
    • 2 years ago
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    Looks correct.

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