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Kinematics ball throwing question

  1. Oct 11, 2006 #1
    hi im having some trouble with the question below, was hoping maybe someone could talk me through it?

    A ball is thrown from a point on an incline of 16 degrees with an initial velocity of 19m/s and at an angle of 49 degrees to the incline. How far up the plane will the ball strike measured from its initial position?

    so far i have resolved the velocity into its x and y directions

    Ux = 8.03m/s
    Uy = 17.22m/s

    So - Sy = Uy*t-(0.5gt^2)
    Sx = Ux*t

    Now i get stuck with the rest, i have to:

    Express the height of the incline, Sp in terms of time, t

    Sp = ____ xt

    At the point of impact, y = Sp, find t:

    t = ____ s

    Now find Sx and Sp

    Sx = ____ m
    Sp = ____ m

    Therefore, the distance up the incline from the point where the ball was released is:

    Sball = _____ m


    i would appreciate any help people can offer me with this as i understand its not that difficult but just cant grasp it.

    thanks les.
     
  2. jcsd
  3. Oct 11, 2006 #2

    berkeman

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    Staff: Mentor

    I don't think I'd use time in this problem. Just write equations for the two lines and set them equal to find the impact point. Write an equation for the y=f(x) of the ball as it flies along, and the y=f(x) straight line of the incline.
     
  4. Oct 11, 2006 #3

    radou

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    Homework Helper

    Another way: let the time of impact equal [tex]t_{0}[/tex]. Then, [tex]y(t_{0}) = v_{0} \sin(65)t_{0} - \frac{1}{2}gt_{0}^2[/tex]. Further on, from the geometry of the incline, you can conclude [tex]y(t_{0})=x(t_{0})\tan(16) = v_{0}\cos(65)t_{0} \tan(16)[/tex]. Solve for [tex]t_{0}[/tex], the rest is trivial.
     
  5. Oct 11, 2006 #4
    thanks for your input so far mate, the only problem is i have to answer all the parts above, basically do it in the way they have set it out.
     
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