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Math Help - Range of a Projectile

  1. #1
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    Range of a Projectile

    ok so I'm having some problems with this problem, it is an optimization problem and the paper says this: The range R of a projectile whose muzzle velocity in meters per second is v, and whose angle of elevation in radians is theta, is given by R=(v^2*sin(2*theta))/g, where g is the acceleration due to gravity. Which angle of elevation gives the maximum range of the projectile?
    if i could get some help figuring out how to do this, it would help me out a lot, thanks
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    Quote Originally Posted by touille89 View Post
    ok so I'm having some problems with this problem, it is an optimization problem and the paper says this: The range R of a projectile whose muzzle velocity in meters per second is v, and whose angle of elevation in radians is theta, is given by R=(v^2*sin(2*theta))/g, where g is the acceleration due to gravity. Which angle of elevation gives the maximum range of the projectile?
    if i could get some help figuring out how to do this, it would help me out a lot, thanks
    find \frac{dR}{d\theta} , set it equal to 0 and maximize.

    remember to treat v and g as constants.
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    so I've got the derivative now and I've set it equal to zero, but the way that i went about solving for zero i don't think it worked, i checked by putting some numbers in for v and g and it didn't give me a zero, i used the double angle formula for cosine because that was the only way that i could figure out how to do it. The derivative that i have gotten is (v^2*cos(2*theta))/g
    Last edited by touille89; March 15th 2010 at 05:03 PM. Reason: needed better wording
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  4. #4
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    Quote Originally Posted by touille89 View Post
    so I've got the derivative now and I've set it equal to zero, but the way that i went about solving for zero i don't think it worked, i checked by putting some numbers in for v and g and it didn't give me a zero, i used the double angle formula for cosine because that was the only way that i could figure out how to do it. The derivative that i have gotten is (v^2*cos(2*theta))/g
    R = \frac{v^2 \sin(2\theta)}{g}

    \frac{dR}{d\theta} = \frac{2v^2\cos(2\theta)}{g}

    \frac{dR}{d\theta} = 0 when \cos(2\theta) = 0


    ... note that \cos(90^\circ) = 0
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    wow, i can't believe i forgot that, thanks so much for helping me out with this, i figured it out, thanks!!!
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