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Math Help - Projectile Motion- Fixed initial velocity show the particle cant move beyond a curve

  1. #1
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    Projectile Motion- Fixed initial velocity show the particle cant move beyond a curve

    A boy throws a ball at a constant speed V for every throw, show that no matter what angle alpha to the horizontal this boy throws the ball at, the ball cannot move beyond a certain curve in the xy plane. Find the equation of this curve. Model the ball as a particle and forget about air resistance.

    This problem has been annoying me for a while now, I believe I have correctly derived all the equations of motion I can, I have the horizontal distance and the vertical distance, I eliminated time from both of these to obtain the path of the ball for a certain angle. I also have the maximum horizontal and vertical distance obtained for a certain angle. I thought to find the maximum distance from the boy that the ball will obtain I should square the horizontal and vertical distances, sum them, differentiate them with respect to t, solve for t for 0 and then substitute t back in, but that didn't get me very far.
    Any ideas?

    Edit: Apparently the answer is y=(V^3 - g^2x^2)/(2gV) but I can't see how to get there
    Last edited by kevinlightman; December 29th 2009 at 04:12 AM.
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  2. #2
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    Hello kevinlightman
    Quote Originally Posted by kevinlightman View Post
    A boy throws a ball at a constant speed V for every throw, show that no matter what angle alpha to the horizontal this boy throws the ball at, the ball cannot move beyond a certain curve in the xy plane. Find the equation of this curve. Model the ball as a particle and forget about air resistance.

    This problem has been annoying me for a while now, I believe I have correctly derived all the equations of motion I can, I have the horizontal distance and the vertical distance, I eliminated time from both of these to obtain the path of the ball for a certain angle. I also have the maximum horizontal and vertical distance obtained for a certain angle. I thought to find the maximum distance from the boy that the ball will obtain I should square the horizontal and vertical distances, sum them, differentiate them with respect to t, solve for t for 0 and then substitute t back in, but that didn't get me very far.
    Any ideas?

    Edit: Apparently the answer is y=(V^3 - g^2x^2)/(2gV) but I can't see how to get there
    I have a solution, but it doesn't quite agree with the one you've been given. In fact, I don't see how that solution can be correct. If we check the dimensions:
    V is LT^{-1} (i.e. speed is length / time)

    g is LT^{-2}

    x is L
    So:
    V^3 is L^3T^{-3}
    and
    g^2x^2 is L^2T^{-4}L^2=L^4T^{-4}
    So the expression V^3-g^2x^2 is dimensionally incorrect.

    Here's my solution.

    Eliminating t between the equations
    x = V\cos\alpha t
    and
    y = V\sin\alpha t - \tfrac12gt^2
    we get
    y = x\tan\alpha - \tfrac12 g \frac{x^2}{V^2}\sec^2\alpha

    \Rightarrow 2V^2y =2V^2x\tan\alpha-gx^2(1+\tan^2\alpha), using \sec^2\alpha = 1+\tan^2\alpha
    Re-arrange as a quadratic in \tan\alpha:
    gx^2\tan^2\alpha - 2V^2x\tan\alpha +2V^2y +gx^2=0
    Now the values of \tan\alpha that satisfy this quadratic must be real. So:
    (-2V^2x)^2-4gx^2(2V^2y+gx^2)\ge0

    \Rightarrow 4V^4x^2 \ge 4gx^2(2V^2y+gx^2)
    Divide by 4x^2, which we may do since x^2\ge 0:
    \Rightarrow V^4 \ge 2gV^2y + g^2x^2

    \Rightarrow y \le \frac{V^4-g^2x^2}{2gV^2}
    Thus the values of x and y are bounded by the curve
    \Rightarrow y = \frac{V^4-g^2x^2}{2gV^2}
    which, I think, is the correct answer. (Unless you can see where I've made a slip?)

    Grandad
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  3. #3
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    Brilliant, I never thought to use the condition that the solutions must be real to introduce an inequality, thanks for the help.
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