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Two Projectile Motion Problems

  1. Jun 3, 2006 #1
    I'll post these in two messages to make this easier. First problem:
    I'm pretty stuck on this one. I've tried juggling all the kinematics equations but the best I can do is solve x=v*cos(theta)*t for t and drop it into y=(1/2)gt^2 + v*sin(theta)*t. This yields the standard Range equation, but I can't solve that for theta. I'm just not sure how to derive the right equation.
     
  2. jcsd
  3. Jun 3, 2006 #2
    Never mind the second problem, I figured it out as I was typing it. Thanks in advance.
     
  4. Jun 3, 2006 #3
    First off, gravity counters the vertical motion of the cannon ball. Your y-equation should thus be y(t)=-(1/2)gt^2+v*sin(theta)*t

    To find theta you will need to determine how long it will take the ball to x(t)=2000m. Once you have found 't' you can then substitute it into your y-equation and solve for theta.
     
  5. Jun 3, 2006 #4
    Yes, but to know how long it takes the ball to travel 2000m along the ground I need to know the x-component of the initial velocity. But I can't get the x-component without theta, which is what I'm trying to find.
     
  6. Jun 3, 2006 #5
    You could always combine the x and y component equations (eliminating t in the process), and solve the resulting equation for the angle, it would probably involve more work, but it would work best overall.
     
  7. Jun 3, 2006 #6
    As stated above, I've tried that. Solving x=vi*cos(theta)t for t and putting into y=(1/2)gt^2 + v*sin(theta)*t yields the standard projectile motion range equation, which is [deep breath]:

    y = tan(theta)*x - [(g*x^2)/(2*vi^2*cos^2(theta))]

    If you can solve that for theta, I will bow to your awesome trig skills.
     
  8. Jun 6, 2006 #7

    siddharth

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    Try using

    [tex]
    \frac{1}{\cos^2 \theta} = sec^2 \theta [/tex]

    [tex]sec^2\theta = 1 + \tan^2 \theta [/tex]

    and solve for [itex] \tan \theta [/itex].
     
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