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Projectile motion

  1. Sep 11, 2011 #1
    1. The problem statement, all variables and given/known data
    A cannon has a muzzle speed of 1000 m/s, and it is used to start an avalanche on a mountain 1000 m from the cannon horizontally and 500 m above the cannon. What angle above the horizontal should the cannon be fired?

    [tex]v_0 = 1000 m/s[/tex]
    [tex]y_t = 500 m[/tex]
    [tex]x_t = 1000 m[/tex]


    2. Relevant equations
    [tex]x_t = x_0 + v_0 cos \theta t[/tex]
    [tex]y_t = y_0 + v_0 sin \theta t - g t^2[/tex]


    3. The attempt at a solution

    I don't know if this approach is right but I tried a system of equations:

    [tex]1000 = 1000 cos \theta t[/tex]
    [tex]1 = cos \theta t[/tex]
    [tex]t = \frac{1}{cos \theta} [/tex]

    I tried substituting t from the first equation into this one:
    [tex]500 = 1000 sin \theta t - (1/2)(9.81)t^2[/tex]
    [tex]500 = 1000 sin \theta t - (1/2)(9.81)t^2[/tex]
    [tex]500 = 1000 \frac {sin \theta }{cos \theta} - \frac{(4.905)}{cos ^2 \theta}[/tex]
    [tex]500 = 1000 tan \theta - (4.905)sec \theta[/tex]

    But I didn't know how to solve for theta. Is this the right approach, or is there an easier way to do this problem?

    [tex][/tex]
     
  2. jcsd
  3. Sep 11, 2011 #2

    vela

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    You should get in the habit of plugging numbers in only near the end. Using a trig identity, you get:
    \begin{align*}
    y_t &= x_t \tan\theta - x_t^2\left(\frac{g}{2v_0^2}\right)\sec^2 \theta \\
    &= x_t \tan\theta - x_t^2\left(\frac{g}{2v_0^2}\right)(1+\tan^2 \theta)
    \end{align*}
    That's quadratic in tan θ. (I'd plug the numbers in now.) Can you take it from here?
     
  4. Sep 12, 2011 #3
    Plugging in from that point I got:

    [tex]500 = 1000 tan \theta - \frac {1000^2}{2*1000^2}*9.81(1+tan^2 \theta)[/tex]

    Which simplified to:

    [tex] tan^2 \theta - 203.874 tan \theta + 102.937 = 0 [/tex]

    I get stuck here. Not sure how to solve this... my idea was let [tex] x = tan \theta [/tex] and solve x using the quadratic formula, then plug [tex] tan \theta [/tex] back in for x to find the angle. Does this work?
    [tex] [/tex]
     
  5. Sep 12, 2011 #4

    vela

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    Yes, that'll work.
     
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