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Projectile trajectories

  1. Dec 30, 2013 #1
    1. The problem statement, all variables and given/known data

    A construction worker stands in a 2.6m deep hole, 3.1m from the edge of the hole. He tosses the hammer to a person outside the hole. If the hammer leaves his hand 1m above the bottom of the hole at an angle of 35, what is the minimum speed needed to clear the edge.

    2. Relevant equations



    3. The attempt at a solution

    Capture.JPG

    Why does the book states the answer = 11m/s?

    This is really frustrating!
    I am getting 7.8......
     
    Last edited: Dec 30, 2013
  2. jcsd
  3. Dec 30, 2013 #2

    SteamKing

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    Draw a sketch of the problem. This may help you setting up the equations of motion. You should be able to determine the book's solution.

    BTW, you can't arbitrarily ignore the negative sign under the square root.
     
  4. Dec 30, 2013 #3

    mfb

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    35° relative to the horizontal?
    Where does your formula come from? It gives a wrong result.

    @SteamKing: the sign is fine, the subtraction in the denominator cancels it.
     
  5. Dec 30, 2013 #4
    The book has given the sketch.

    In fact, the book has given the equation(in symbolic expression) and stated the answer is 11m/s but upon substituting the x and y coordinatate of 3.1m and 1.6m respectively, I don't arrive at 11m/s.
    The book has given a sketch too.
     
  6. Dec 30, 2013 #5
    Capture.JPG
     
  7. Dec 30, 2013 #6
    It comes from t = x/vi cos theta

    y = vi sin theta . t - 0.5g t^2

    sub t into y-coordinate to get y as a function of x-coordinate.
    Once the equation is derived, sub x= 3.1 and y = 1.6
    but it doesn't give me 11m/s if I express it in terms of vi
     
  8. Dec 30, 2013 #7

    mfb

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    I can confirm the book's answer. You did something wrong at solving the equation for the velocity.
     
  9. Dec 30, 2013 #8

    SteamKing

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    I suggest you check your calculations again.

    (3.1 tan 35) - 1.6 > 0
     
  10. Dec 30, 2013 #9

    mfb

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    Oh sorry, for some reason I squared the tan instead of the cos.
     
  11. Dec 30, 2013 #10
    Still not getting the answer.

    I basically run the calculation of the workings in my OP but it doesn't turn up to be 11m/s
     
  12. Dec 30, 2013 #11

    mfb

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    The approach of post #6 is right, the formula in post #1 is not. If you need help to find your error, show the steps you made in between.
     
  13. Dec 30, 2013 #12
    Post 1 is nothing more than a mathematical expression of post 6.

    I'm sure post 1 is right. I think the issue might be that on my calculator I did (2cos35)^2 instead of 2cos(35)^2
     
  14. Dec 30, 2013 #13
    I'm getting 10.036m/s
     
  15. Dec 30, 2013 #14
    x = vcos(theta).t
    y = vsin(theta).t - 0.5gt^2

    t = x/(vcos(theta))

    y = vsin(theta)[(x)/vcos(theta)] - 0.5g[(x)/vcos(theta)]^2

    y = xtan(theta) - [(g)/(2v^2 (cos(35)^2))

    v = SQRT[(gx^2)/(2cos(theta)^2(xtan(theta)-y))

    Sub x = 3.1 and y = 1.6
    g = -9.8

    It's giving me 10m/s and not 11m/s
     
  16. Dec 30, 2013 #15

    mfb

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    You are using the wrong sign for g. Use a negative sign in the equation for y=... or for g itself, but not both, otherwise you are accelerating upwards.

    That cannot be the only error (check -> 12.5), but I don't see the error right now.
     
  17. Dec 30, 2013 #16

    I don't see any other error aside from the negative g that I use. And why shouldn't g be negative? It has always been negative according to convention
     
  18. Dec 30, 2013 #17

    Besides, you're not getting 11m/s too. The link from you shows wolfram getting 12m/s+
     
  19. Dec 30, 2013 #18

    Check the jpg in my op. -9.8 in the working indicates a negative g.
     
  20. Dec 31, 2013 #19
    I have found v to be 11m/s.

    To find where the hammer lands:

    Capture.JPG

    This is a very juvenile question to ask but is the a, b and c values correct?
    Solving via quadratic, I'm not getting x= 3.1 or x = 8.7.
    It's a very simple problem but this question is taking me more than 1 day to solve!!!!
     
  21. Dec 31, 2013 #20

    mfb

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    Okay, but if g is negative (so y is pointing upwards), the equation of motion is ##y=v_y t + \frac{1}{2} g t^2## and not minus.

    That's why I said "this cannot be the only error". Check the WolframAlpha link in post 7, 11m/s is right.

    What do you get for x with the reverse attempt? Keep in mind that 11m/s is not the exact value.
     
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