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Optimum Angle of Launch

  1. Jan 20, 2007 #1
    I would like to know what the Optimum Angle of Launch would be for an object projecting forward to obtain maximum distance.

    The object will be traveling approximately 25 mph and will launch off a surface 2 feet above the ground.

    What is the Optimum Angle of Launch to get that object to the farthest point on the ground?

    Thanks for any help...
     
  2. jcsd
  3. Jan 20, 2007 #2
    I was told that 45 degrees is optimum on a level horizontal plane. In other words the if the object was projected from ground.

    However, in this case the object is projected from 2 feet above the ground.

    Does that make a difference?
     
  4. Jan 20, 2007 #3
    I believe in this case the angle is slightly less than 45 degrees.

    you can work it out by brute forcing. find the time it takes to land, then substitute t in the x equation, then differentiate to find maximum R.
     
  5. Jan 21, 2007 #4

    D H

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    The optimal angle is indeed less than 45 degrees if the object starts its trajectory above the ground. An example: Suppose the object is thrown 2 feet above the ground with an initial speed of 8 .05 ft/second. A 45 degree angle results in the object going 3.24 feet. The distance is 3.46 feet at the optimal angle of 30 degrees.

    This looks like homework. The original poster should show some work.
     
    Last edited by a moderator: Jan 21, 2007
  6. Jan 21, 2007 #5

    arildno

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    The optimal launch angle does, indeed, depend on the launching position and the ground profile.

    Assume that the initial vertical position, launch speed, launch angle, ground profile is [itex]y_{0}, v, \theta, h(x)[/itex] respectively.
    Then, as a function of time, the object's vertical and horizontal positions (starting at x=0) are
    [tex]y(t)=y_{0}+(v\sin\theta){t}+\frac{gt^{2}}{2}, x(t)=(v\cos\theta){t}[/tex]
    The object hits the ground at some time T when y(T)=h(x(T)), that is we gain the ground state equation:
    [tex]h((v\cos\theta)T)=y_{0}+(v\sin\theta)T+\frac{gT^{2}}{2}[/tex]
    an equation by which we in principle can solve for the collision instant T as a function of [itex]\theta[/itex], called [itex]T(\theta)[/itex]

    The RANGE is therefore the horizontal coordinate considered as a function of the launch angle:
    [tex]x(\theta)=v\cos\theta{T}(\theta)[/tex]
    and the optimal launch angle [itex]\theta_{op}[/itex] is determined by solving the the algebraic equation [tex]\frac{dx}{d\theta}=0[/tex], that is finding the solutions of:
    [tex]-(\sin\theta_{op})T(\theta_{op})+(\cos\theta_{op})\frac{dT}{d\theta}\mid_{(\theta=\theta_{op})}=0\to\frac{1}{T(\theta_{op})}\frac{dT}{d\theta}\mid_{(\theta=\theta_{op})}=\tan(\theta_{op})[/tex]

    In the general case, it is by no means a trivial matter to determine the function [itex]T(\theta)[/itex], nor is it trivial to solve the range equation for [itex]\theta_{op}[/itex]
     
    Last edited by a moderator: Jan 21, 2007
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