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anonymous

  • one year ago

"A novice golfer on the green takes three strokes to sink the ball. The successive displacements of the ball are d1 = 4.00 m to the north, d2 = 1.90 m northeast, and d3 = 1.10 m at θ = 25.0° west of south (figure below). Starting at the same initial point, an expert golfer could make the hole in what single displacement? " I can't seem to understand why the answers I'm getting are wrong. I found the displacement between d2 and d3 to be 1.54919m and used that as the third leg for d1, making R to be 4^2 + 1.54919^2, and got 4.29 as the answer. This was wrong but within 10%, but I don't see

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  1. anonymous
    • one year ago
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    Can I see the figure to make sure?

  2. anonymous
    • one year ago
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    Problem attached via screenshot. Thank you.

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  3. anonymous
    • one year ago
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    You add all the x and y component together then you would use Pythagorean theorem \[total x component^2+total y component^2=displacement^2\]

  4. anonymous
    • one year ago
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    I did this. Got 4.43m for the answer, which I think is correct. Haven't checked yet, I'm on my last answer. And the angle I got was 78.57 degrees, but again unsure of that's correct.

  5. anonymous
    • one year ago
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    X component Y component D1 0m 4m D2 1.90cos45 1.90sin45 D3 -1.10sin25 -1.10cos25 For D3 it would be negative on both side since they are moving down and to the left. For D2 it would be a 45 degree angle since it did not indicate each angle. |dw:1442814882099:dw|

  6. anonymous
    • one year ago
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    1.10 is hypotenuse* Add all up of x and y component separately and use Pythagorean theorem to get displacement.

  7. anonymous
    • one year ago
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    What I did was go 4(cos(90)+1.9(cos45)+1.1(cos245)* = .878622796 4sin(90)+1.9cos(45)+1.1(cos245)= 4.3465 X = .878622796 y = 4.3465 H = 4.43m A = tan y/x = 78.57degrees

  8. anonymous
    • one year ago
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    245 because it's 25 to 270.

  9. anonymous
    • one year ago
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    Hmm, I do not use angles greater than 90 degrees on this kind of problems I use only from 0 to 90 degrees. If it is going west or south, I just make it negative. If it is going north or east I just leave it as positive

  10. anonymous
    • one year ago
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    Wouldn't it be beyond those bounds regardless? Either way, are those correct? I don't think I'm allowed any more mistakes.

  11. anonymous
    • one year ago
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    My way was correct on both. Your way resulted in the same answers. Thank you for the help, though. My physics professor has a habit of giving us questions he didn't remotely teach us how to solve in class and won't give any help outside of that. He even said to come in and ask him after class the day after it was due. Either way, thank you. Best answer!

  12. anonymous
    • one year ago
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    Oh ye, my mistake. The way you put cos245 on both components confused me. But I see how you figured it out. Good luck with the rest!

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