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Ball Motion Down the Incline

  1. Mar 27, 2009 #1
    Hi. I have problem with my homework. I have already searched related theoretic basis and I have already tried to solve it too, but I need some feedback, if my solution is right. I would like to ask you for it. Thank you very much.

    1. The problem statement, all variables and given/known data
    http://katus.kabel1.cz/homework.jpg [Broken]
    The ball falls down on the incline from height h (known variable). Then it rebounds according to the law of reflection (angle δ). The ball has to fly through the hole in the wall standing in distance x (known variable). Angle of the incline is α.
    Given data:
    [tex]h = 0,5 m[/tex]
    [tex]x = 0,15 m[/tex]
    [tex]\alpha = 15°[/tex]
    and all the variables in the picture typed bold.

    2+3. Relevant equations and the attempt at a solution
    Free fall
    acceleration:
    [tex]a = g[/tex]
    velocity:
    [tex]v = \int a \ dt = g t + v_0 [/tex]
    height:
    [tex]h = \int v \ dt = \frac{1}{2} g t^2 + v_0 t [/tex]
    [tex]v_0 = 0[/tex]
    time of impact:
    [tex]t = \sqrt {\frac{2h}{g}}[/tex]
    velocity in the time of impact:
    [tex]v = g t = \sqrt {2gh}[/tex]

    Oblique throw
    accelerations:
    [tex]a_x = 0[/tex]
    [tex]a_y = -g[/tex]
    velocities:
    [tex]v_0 = \sqrt {2gh}[/tex]
    [tex]v_x = \int a_x \ dt = v_0 cos \beta [/tex]
    [tex]v_y = \int a_y \ dt = -g t + v_0 sin \beta [/tex]
    distances:
    [tex]x = \int v_x \ dt = v_0 cos \beta t + x_0[/tex]
    [tex]y = \int v_y \ dt = - \frac{1}{2} g t^2 + v_0 sin \beta t + y_0[/tex]
    [tex]x_0 = 0[/tex]
    [tex]y_0 = 0 [/tex]
    time of contact with the wall:
    [tex]t = \frac{x}{v_0 cos \beta}[/tex]
    [tex]y = x tg \beta - \frac{x^2}{4 h (cos \beta)^2}[/tex]

    My results with the given data:
    [tex]y = 0,257 m[/tex]
    Is it right, please?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 27, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Welcome to PF.

    It seems a little complicated to not solve for the intermediate results along the way, but your answer looks in close agreement to what I get.

    I think your last statement for y is not quite right, however.
     
  4. Mar 28, 2009 #3
    Thank you very much for your help!

    Do you think the last y statement is wrong?

    I get it from the previous statements this way:

    [tex]x = v_0 cos \beta \cdot t[/tex]
    [tex]t = \frac {x}{v_0 cos \beta}[/tex]
    [tex]y = - \frac {1}{2}g t^2 + v_0 sin \beta \cdot t[/tex]
    [tex]y = - \frac {1}{2}g (\frac {x}{v_0 cos \beta})^2 + v_0 sin \beta \cdot {\frac {x}{v_0 cos \beta}}[/tex]
    [tex]y = - \frac {1}{2}g \frac {x^2}{v_0^2 (cos \beta)^2} + sin \beta \cdot {\frac {x}{ cos \beta}}[/tex]
    [tex]v_0 = \sqrt{2gh}[/tex]
    [tex]y = - \frac {1}{2}g \frac {x^2}{2gh (cos \beta)^2} + sin \beta \cdot {\frac {x}{cos \beta}}[/tex]
    [tex]y = - \frac {1}{4} \frac {x^2}{h (cos \beta)^2} + x \cdot {\frac {sin \beta}{cos \beta}}[/tex]
    [tex]y = - \frac {x^2}{4 h (cos \beta)^2} + x \cdot tg \beta[/tex]
     
  5. Mar 28, 2009 #4

    LowlyPion

    User Avatar
    Homework Helper

    Sorry. I didn't recognize your tgβ notation as tanβ with t and g also being variables of the solution.
     
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