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Homework Help: Golf ball problem (velocity, max height, hang time)

  1. Oct 18, 2006 #1
    "golf ball problem" (velocity, max height, hang time)

    Okay. I'm having a really hard time with this. I know how to start the problem, but after that I go nuts, and literally start pulling out my hair. There is no book I can look in for reference, as this part is teacher supplemented.

    The given picture/information is this:

    physics-question.jpg

    And I need to find:

    velocity of x at zero time:
    velocity of y at zero time:
    maximum height (from ground, not cliff):
    hang time:
    velocity of x after "a" time:
    velocity of y after "a" time:

    Now, I can find Vx at zero (20m/s), and Vy at zero (34.6m/s). I also know that Vx at any time is going to remain the same (20m/s), but then I'm lost for maximum height, hang time, and Vy at "a" time. Gravity is taken into account, but not air resistance. I know that it loses -9.8m/s going up due to gravity, and +9.8m/s going down due to gravity. And I tried doing that a long way, subtracting 9.8 from 34.6, subtracting 9.8 form that answer, etc., and counting how many times to get a seconds number, but then I didn't know how to figure out when I got down to 5.2 and subtracting 9.8 from that. I really have no clue how far up it goes, or how to find anything past making the triangle to find the forces using sine/cosine/tangent.

    I'm not sure if there is a formula or multiple formulas to figure this information out. That's what I think I'm missing to get the rest of the information.
     
  2. jcsd
  3. Oct 18, 2006 #2

    radou

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    Homework Helper

    The formula which will be of great use to you is the displacement formula. For the x direction, you have [tex]x(t) = x_{0} + v_{0x}t[/tex], and, for the y direction, [tex]y(t) = y_{0} + v_{0y}t - \frac{1}{2}gt^2[/tex]. You should be able to solve the problem now.
     
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