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Math Help - simple, challenging catapult problem

  1. #1
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    simple, challenging catapult problem


    Above is a catapult arm with it's rope hugging it. It is at 85degrees with the vertical.

    The catapult arm was moved 60degrees to the opposite direction, released, and fired at a range of 35m.
    Considering everything else is constant, what is the rate of change(dr/dΘ) of range with respect to how much angle the spoon was moved?
    Last edited by markxchan; February 9th 2010 at 12:58 PM.
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  2. #2
    zeg
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    Suggestion

    The equation for the range of an object is:
    r = \frac{v_isin(2\theta_i)}{g}
    where v_i is the initial velocity, \theta_i is the initial angle of trajectory with respect to the horizontal, and g is acceleration due to gravit, 9.81 m/s^2. I would suggest finding \theta_i in terms of spoon angle, \theta_s and then take your \frac{dr}{d\theta_s}.
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  3. #3
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    Quote Originally Posted by zeg View Post
    The equation for the range of an object is:
    r = \frac{v_isin(2\theta_i)}{g}
    where v_i is the initial velocity, \theta_i is the initial angle of trajectory with respect to the horizontal, and g is acceleration due to gravit, 9.81 m/s^2. I would suggest finding \theta_i in terms of spoon angle, \theta_s and then take your \frac{dr}{d\theta_s}.
    That's what I thought too.

    It is actually more complicated than that as V is not constant. V changes depending on the force the spoon/rope was pulled and you can know how much force is on the rope because it increases as angle gets closer to 0.

    I think what complicates this problem for me is that I need to calculate the force exerted by the string on the spoon as it is rotated(by a hand) and released, and I don't know how to do that.
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