I really need help.

- AmTran_Bus

I really need help.

- chestercat

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- 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

If I have it drawn right, part a is the pyth. theorem.

- AmTran_Bus

|dw:1360440654830:dw|

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## More answers

- AmTran_Bus

Maybe law of sines for part two?

- klimenkov

I am not very good in English, but if your pic is correct, you found the distance right.

- AmTran_Bus

Thanks so much! I think I drew it right. @Jonask what do you think?

- anonymous

yes for 2|dw:1360448299629:dw|

- AmTran_Bus

Awesome. Could I take the arctan?

- AmTran_Bus

Whoops, no. I ment the law of sines.

- klimenkov

\(\theta\) on the @Jonask 's pic is not measured from the horizon.

- 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

Where is the horizon?

- AmTran_Bus

I wonder if it is a right triangle, from part 4.

- klimenkov

|dw:1360441416725:dw|

- AmTran_Bus

Oh boy.

- anonymous

yes its like an angle of elevation @klimenkov |dw:1360448668700:dw|

- AmTran_Bus

yippee. So where do I go from here?

- AmTran_Bus

1st of all, is the pic 100% correct?

- 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

*part 3

- AmTran_Bus

Anybody?

- anonymous

tan theta=3.2/15.3

- anonymous

not theta but alpha\[\tan \alpha=3.2/15.3\]

- AmTran_Bus

That is 0.00365.

- anonymous

now to get alpha use arctan(0.00365)

- AmTran_Bus

That is 0.209129

- AmTran_Bus

Thanks @Jonask

- AmTran_Bus

Now for the next part///

- 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

I was thinking the same, had it wrote down. Wanted to verify. Thanks SOO much.

- AmTran_Bus

So what needed to be done to find the angle from the horizon?

- anonymous

its that arctan 0.00365=0.209 degrees
this also accounts for the balls imppsossible perpetual linear motion

- AmTran_Bus

Thanks and God Bless. Keep on living for Jesus.

- anonymous

God Bless you too @AmTran_Bus

- anonymous

@klimenkov

- klimenkov

Looks correct.

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