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Math Help - Maximizing The Range Of A Projectile

  1. #1
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    Maximizing The Range Of A Projectile

    The range of a projectile is R = v0^2*sin(2*theta)/g where v0 is its initial velocity, g is the acceleration due to gravity and is a constant, and theta is its firing angle. Find the angle that maximizes the projectile's range.

    I know the answer is 45 degrees, but I'm confused as to how to arrive at that answer using optimization.
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  2. #2
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    Quote Originally Posted by Fred119 View Post
    The range of a projectile is R = v0^2*sin(2*theta)/g where v0 is its initial velocity, g is the acceleration due to gravity and is a constant, and theta is its firing angle. Find the angle that maximizes the projectile's range.

    I know the answer is 45 degrees, but I'm confused as to how to arrive at that answer using optimization.

    Derivate the function R(\theta)=\frac{v_0^2\sin(2\theta)}{2} wrt \theta and equate the derivative to zero:

    R_{\theta}=v_0^2\cos(2\theta)=0\,\Longrightarrow\,  2\theta=\pi \,\,radians=90\,\, degrees\, \Longrightarrow\,\theta=\frac{\pi}{2}\, radians=45\,\,degrees

    Tonio
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    Quote Originally Posted by tonio View Post
    Derivate the function R(\theta)=\frac{v_0^2\sin(2\theta)}{2} wrt \theta and equate the derivative to zero:

    R_{\theta}=v_0^2\cos(2\theta)=0\,\Longrightarrow\,  2\theta=\pi \,\,radians=90\,\, degrees\, \Longrightarrow\,\theta=\frac{\pi}{2}\, radians=45\,\,degrees

    Tonio
    Why did you rewrite the function over 2 instead of g? Thank you.
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  4. #4
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    Quote Originally Posted by Fred119 View Post
    Why did you rewrite the function over 2 instead of g? Thank you.

    My bad, as simple as that...but the derivative vanishes in the very same point as before, so nothing REALLY important changes in the proof.

    Tonio
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