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Math Help - [SOLVED] Physics - Projectile Motion (2d), Vectors

  1. #1
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    [SOLVED] Physics - Projectile Motion (2d), Vectors

    Problem: A projectile is fired in such a way that its horizontal range is equal to 14 times its maximum height. What is the angle of projection?


    The work I have done thus far:

    Horizontal range R = 14h, where h is the maximum height.

    The Cartesian coordinates for the maximum height are given by (R/2, h). So, since R/2 = (14h/2) =7h, the coordinates are (7h, h).

    This gives a right triangle with height h and base R/2 = 7h

    Thus, tan(theta) = (h / 7h) = 1/7

    So, arctan(1/7) = theta = 8.13 degrees approximately

    This is basically where I'm at. I don't know what to do now or even if I'm heading in the right direction.



    P.S. is there a sticky thread or web page that shows a list/tutorial on using the math codes/formatting. The only format I know how to use are the [ math ] tags but they are not always working intuitively for me.

    For example, I had to use arctan instead of tan^-1 because the tags did not format ^-1 as in superscript as an exponent would beI know it's not an exponent in this case, it signifies the "inverse of", but you get my drift. Also, I do not know how to display the "degrees" symbol or Greek letters. Is there a way to do this.

    I would list relevant formulas for R and h, but I wouldn't know how to format them. Most of them can be found here.

    Thanks.
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  2. #2
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    Hi

    I'm afraid you are wrong

    A projectile fired with an angle \alpha to the horizontal and with his weight as only force acting on it describes a parabola

    You can start from the equation of the acceleration
    m \vec{a} = m \vec{g}

    Choosing an appropriate system of axes (O,x,y) O being the point where the projectile is fired, (Ox) being the horizontal directed in the same direction of the motion

    a_x = 0
    a_y = -g

    Integrating gives the speed
    v_x = v_0 \cos \alpha
    v_y = -gt + v_0 \sin \alpha

    \vec{v_0} being the initial speed

    Integrating again gives the coordinates from which you can find the range and the maximum height and finally \alpha

    You will find some help on LATEX here
    http://www.mathhelpforum.com/math-help/latex-help/
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  3. #3
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    Quote Originally Posted by lightningstab714 View Post
    Problem: A projectile is fired in such a way that its horizontal range is equal to 14 times its maximum height. What is the angle of projection?
    both of the following equations can be derived from the basic kinematics equations of motion with constant acceleration ...

    h = \frac{(v_0 \sin{\theta})^2}{2g}

    R = \frac{2v_0^2 \sin{\theta}\cos{\theta}}{g}<br />


    \frac{2v_0^2 \sin{\theta}\cos{\theta}}{g} = 14 \frac{(v_0 \sin{\theta})^2}{2g}

    2\cos{\theta} = 7 \sin{\theta}

    \tan{\theta} = \frac{2}{7}

    \theta = \arctan\left(\frac{2}{7}\right) \approx 16^\circ
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  4. #4
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    Thanks to both of you.
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