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Kinematics - Projectile

  1. Feb 18, 2014 #1
    1. The problem statement, all variables and given/known data

    A stone is thrown with a velocity [itex]v_{0}[/itex] at an angle [itex]\alpha[/itex] to the horizontal (see image) from a step of height H. Calculate the x coordinate [itex]x_{1}[/itex] of the point where the stone hits the ground.

    2. Relevant equations


    [itex]x(t) = u_{x}t + x_{0}[/itex]
    [itex]y(t) = u_{y}t + \frac{1}{2}at^{2} + y_{0}[/itex]

    3. The attempt at a solution

    [itex]u_{x} = v_{0}cos(\alpha)[/itex]
    [itex]u_{y} = -v_{0}sin(\alpha)[/itex]

    [itex]x_{1} = u_{x}t_{1}[/itex]
    [itex]x_{1} = v_{0}cos(\alpha)t_{1}[/itex]
    [itex]y(t) = 0 @ t_{1}[/itex]
    [itex]0 = -v_{0}sin(\alpha)t_{1} - \frac{g}{2}(t_{1})^2 + H[/itex]


    Here is where I have a discrepency with the provided (Bare Bones) Solution.

    According to the solution; [itex]0 = v_{0}sin(\alpha)t_{1} - \frac{g}{2}(t_{1})^2 + H[/itex]

    This seems to be missing the "-" in front of the 0 = [itex]v_{0}sin(\alpha)t_{1}[/itex] term.

    My question is am I correct?
     

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  2. jcsd
  3. Feb 18, 2014 #2
    You could say that both your result and the solution are correct; they just make different assumptions about how the angle α is determined.

    Your result assumes that α is a positive angle which lies between the positive x-axis and the initial velocity vector.

    For the given solution to be correct, it is assumed that the angle α must be determined by rotating counter-clockwise from the positive x-axis for positive angles (or clockwise for negative angles).
     
  4. Feb 18, 2014 #3
    [tex]x(t)=v_0\cos\alpha t, y(t)=H+v_0\sin\alpha t-\frac{1}{2}gt^2[/tex]
    combine those two parametric equations, we get
    [tex]y=x\tan\alpha-\frac{g\sec^2\alpha}{2v_0^2}x^2+H[/tex]

    let y=0 and solve for x
     
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