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Simple trajectory, maximizing downhill range

  1. Dec 5, 2006 #1
    1. The problem statement, all variables and given/known data
    A projectile is fired from the origin down an inclined plane that makes an angle, b, with the horizontal. The angle of elevation of the gun and the initial speed of the projectile are a and V, respectively.

    Show that the angle of elevation, a, that will mazimize the downhill range is the angle halfway between the plane and the vertical.




    2. Relevant equations
    I've already found the standard parametric equations of the trajectory as a function of time, t, below:

    x(t)=(Vcos(a))t and y(t)=(Vsin(a))t - .5gt^2


    3. The attempt at a solution
    I've already found the downhill range to be:

    [(Vcos(a))t]/cos(b)

    and if the angle that maximizes the downhill range is indeed the angle halfway between the plane and the vertical, then:

    (b+(pi/2))/2 = a + b

    a = pi/4 - b/2

    I just have no idea where to go after this, I've been trying to figure out a way to represent the time, t, at which the particle strikes the downhill plane as a function of V, a, and b. All has been to no avail so your guidance is much appreciated!
     
  2. jcsd
  3. Dec 6, 2006 #2

    OlderDan

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    Homework Helper

    You can eliminate the time parameter from your first two equations to find y vs x instead of x(t) and y(t). You can also write an equation for y vs x representing the surface of the ground. The downhill range can be calculated from the coordinates of the intersection of these two expressions for y vs x. How does he downhill range relate to the x coordinate of the intersection. Does maximizing one of these maximize the other?
     
  4. Dec 6, 2006 #3
    Thank you!
     
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