AmTran_Bus
  • AmTran_Bus
I really need help.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
AmTran_Bus
  • AmTran_Bus
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?
AmTran_Bus
  • AmTran_Bus
If I have it drawn right, part a is the pyth. theorem.
AmTran_Bus
  • AmTran_Bus
|dw:1360440654830:dw|

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AmTran_Bus
  • AmTran_Bus
Maybe law of sines for part two?
klimenkov
  • klimenkov
I am not very good in English, but if your pic is correct, you found the distance right.
AmTran_Bus
  • AmTran_Bus
Thanks so much! I think I drew it right. @Jonask what do you think?
anonymous
  • anonymous
yes for 2|dw:1360448299629:dw|
AmTran_Bus
  • AmTran_Bus
Awesome. Could I take the arctan?
AmTran_Bus
  • AmTran_Bus
Whoops, no. I ment the law of sines.
klimenkov
  • klimenkov
\(\theta\) on the @Jonask 's pic is not measured from the horizon.
anonymous
  • anonymous
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.
AmTran_Bus
  • AmTran_Bus
Where is the horizon?
AmTran_Bus
  • AmTran_Bus
I wonder if it is a right triangle, from part 4.
klimenkov
  • klimenkov
|dw:1360441416725:dw|
AmTran_Bus
  • AmTran_Bus
Oh boy.
anonymous
  • anonymous
yes its like an angle of elevation @klimenkov |dw:1360448668700:dw|
AmTran_Bus
  • AmTran_Bus
yippee. So where do I go from here?
AmTran_Bus
  • AmTran_Bus
1st of all, is the pic 100% correct?
AmTran_Bus
  • AmTran_Bus
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.
AmTran_Bus
  • AmTran_Bus
*part 3
AmTran_Bus
  • AmTran_Bus
Anybody?
anonymous
  • anonymous
tan theta=3.2/15.3
anonymous
  • anonymous
not theta but alpha\[\tan \alpha=3.2/15.3\]
AmTran_Bus
  • AmTran_Bus
That is 0.00365.
anonymous
  • anonymous
now to get alpha use arctan(0.00365)
AmTran_Bus
  • AmTran_Bus
That is 0.209129
AmTran_Bus
  • AmTran_Bus
Thanks @Jonask
AmTran_Bus
  • AmTran_Bus
Now for the next part///
anonymous
  • anonymous
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
AmTran_Bus
  • AmTran_Bus
I was thinking the same, had it wrote down. Wanted to verify. Thanks SOO much.
AmTran_Bus
  • AmTran_Bus
So what needed to be done to find the angle from the horizon?
anonymous
  • anonymous
its that arctan 0.00365=0.209 degrees this also accounts for the balls imppsossible perpetual linear motion
AmTran_Bus
  • AmTran_Bus
Thanks and God Bless. Keep on living for Jesus.
anonymous
  • anonymous
God Bless you too @AmTran_Bus
anonymous
  • anonymous
@klimenkov
klimenkov
  • klimenkov
Looks correct.

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