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Projectile Equation - Parabola

  1. Jun 24, 2012 #1
    Here's a equation of a projectile thrown in a parabolic path :

    y=xtanθ - gx2/2u2cos2θ

    where x is the corresponding x - coordinate or the range of a projectile at a point ,
    y is the corresponding y - coordinate or the height of projectile at point,
    θ is the angle made with the horizontal arbitrary x-axis through which projectile is thrown.
    u is magnitude of initial velocity vector making angle θ with horizontal.

    Note: Point from where the projectile began is assumed to be origin , i.e. (0,0) coordinate.

    I was once told that in order to solve difficult problems of projectile , calculus is a very powerful tool.

    We can consider y in equation as a function of θ and then can differentiate y with respect to θ {find y'(θ)}.

    dy/dθ = .........

    Then we can put dy/dθ =0 and also dx/dθ =0. Hence we can easily solve for optimal angle θo for maximum range. Sometimes we just put dy/dθ =0 when we are given "x" and we can solve for θ.

    Also if we imagine the equation to be plotted on a graph then we can obviously consider y as a function of x and then differentiate y with respect to x {find y'(x)}.

    Then we put dy/dx=0 to find point of maxima or minima , I think. We also find d2y/dx2.

    So here are my questions :

    What's the use or when do we find dy/dθ or dy/dx in the projectile equation? I cannot understand. Also I cannot fathom this all theoretically. Why are we doing such thing ? Can someone explain ? Is this really a powerful tool to solve projectile problems ? When do we put dy/dθ =0 and also dx/dθ =0 or even dy/dx =0 and why , while differentiating w.r.t θ or x in y=xtanθ - gx2/2u2cos2θ ?
     
    Last edited: Jun 25, 2012
  2. jcsd
  3. Jun 25, 2012 #2

    haruspex

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    Since x and θ are both being considered as variables here, in each case, it should really be the partial derivative: ∂y/∂θ, ∂y/∂x etc. E.g. ∂y/∂θ says how y changes when θ is varied but x is held constant.
    ∂y/∂x = tanθ - gx/u2cos2θ:
    For a given θ, ∂y/∂x tells you the slope of the trajectory as you look along it. When this becomes zero, the projectile is moving horizontally, so is at the peak of the parabola:
    tanθ = gx/u2cos2θ
    u2 sin(2θ) = 2gx
    I.e. the distance at which the arc peaks is u2 sin(2θ) /2g.

    For max range, we would first find what the range is. This is the nonzero x which makes y zero:
    0 = xrangetanθ - gxrange2/2u2cos2θ
    xrange = u2 sin(2θ)/g
    (twice the distance to the peak, unsurprisingly). To maximise wrt θ:
    0 = dxrange/dθ = 2u2 cos(2θ)/g
    So θ = π/4.

    It's not obvious to me what use would be made of setting ∂x/∂θ = 0 in the original equation. ∂y/∂θ will be 0 when y is at the maximum value it can reach by varying θ only. This value may depend on x:

    ∂(x tanθ - gx2/2u2cos2θ)/∂θ =
    x sec2θ - (gx2/2u2)2 sec2θ tanθ
    Setting this zero:
    1 = (gx/u2) tanθ
    tan θ = u2/gx
    That is, if I want y to be as great as possible at distance xcrit then I should set θ = atan( u2/gxcrit)
     
  4. Jun 26, 2012 #3
    What if y changes when both x and θ are being varied. What will we get when we find ∂y/∂θ or ∂y/∂x or dy/du even in equation ? :

    y=xtanθ - gx2/2u2cos2θ

    Also what if u is varied when y changes but θ and x both are held constant. Logically , I guess this does not even make sense.

    Well ok , but in this case , it'll mean that how y changes with respect to x but θ is held constant. This does not make sense to me.

    Well ok...

    So here this means that y changes w.r.t to θ but x is held constant. What if we do not set ∂y/∂θ =0 ? Also what if we do not set ∂x/∂θ =0 in projectile equation ? What new can we get ?

    Also if in equation :

    y=xtanθ - gx2/2u2cos2θ .... (i)

    I put x = u2sin2θ/g

    I get y = (2u2sin2θ - u4tanθ)/g

    If I put y=u2sin2θ/2g in (i)

    I obtain x= u2sin2θ/2g

    If I put y=u2sin2θ/2g and x = u2sin2θ/g
    in (i) I get :

    u2 = sinθcosθ
    2u2 = sin2θ

    These equations I obtain , are they correct ?

    Moreover you got :

    tan θ = u2/gx
    x= u2/gtanθ
    If I set
    x = u2sin2θ/g

    Then u2sin2θ/g = u2/gtanθ

    I get after solving that θ = ∏/4

    Moreover ,

    These are for symmetric projectile. What if projectile is thrown from an elevation at an angle from horizontal or to an elevation ? What can we get in that case from projectile equation - i.e what will be optimal angle for maximum range ?
     
  5. Jun 26, 2012 #4

    haruspex

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    If you make a small change Δx to x and Δθ to θ then the change in y will be Δy = Δx∂y/∂x + Δθ∂y/∂θ
    Yes, it makes sense. Holding θ constant is obvious; holding x constant at [itex]\hat{x}[/itex] means you're focusing on the point in the trajectory distance [itex]\hat{x}[/itex] from the start. Then you can vary the launch velocity and observe the effect on the height at [itex]\hat{x}[/itex].
    It means watching what happens to the height at a given distance as you vary θ.
    Maybe you don't understand the significance of the places where a derivative becomes zero.
    If we fix x = [itex]\hat{x}[/itex] and vary θ, we see how the height of the trajectory varies at distance [itex]\hat{x}[/itex]. If ∂y/∂θ > 0, increasing θ will increase y. If ∂y/∂θ < 0, increasing θ will decrease y. To get the maximum y, we want to increase θ so long as that makes y increase. As soon as it gets to the point where it would make y decrease we want to stop changing θ. Since these equations involve smooth functions, that change-over cannot happen suddenly. To get from ∂y/∂θ > 0 to ∂y/∂θ < 0 it must pass through the point ∂y/∂θ = 0. Likewise, for a transition from ∂y/∂θ < 0 to ∂y/∂θ > 0.
    So when the value of θ, for the given [itex]\hat{x}[/itex], is such that ∂y/∂θ = 0, we must be at either a local maximum or local minimum of y wrt θ.
    [OK, there is a third option: it could be an inflexion point. This happens when ∂y/∂θ hits zero but instead of crossing to the opposite sign it immediately turns around again: either >0, 0, > 0 or < 0, 0, < 0. So in general it is not enough to observe the derivative is zero at some point: you should look at whether it is positive or negative just either side of that point. In the present case however we are clearly dealing with local maxima.]
    So, to answer your question, if you do not set ∂y/∂θ = 0 you will not be looking at a local extremum. You could set it to something else, but I struggle to think how you would usefully do that in the context of your set-up.
    In the context we're discussing, y is the dependent variable. It is a function of the independent variables x, θ and u. The partial derivatives are only defined on that basis; e.g., it is only by knowing that u and x are the other independent variables that you know to interpret ∂y/∂θ as varying θ while holding x and u fixed. So it doesn't make sense to ask about ∂x/∂θ.
    However, you can recast the equation, making x a function of y, u and θ. Provided you're clear that this is what you are doing, that now gives ∂x/∂θ a meaning. Namely, the recast has given you an equation that tells you at what distance the projectile will have a specified height, given its initial speed and angle. ∂x/∂θ now tells you how that will change as you vary the angle.
    No, you should get y = u22cos2θ/g - u22sin2θ/g = 2u2cos(2θ)/g
    I could tell you'd gone wrong somewhere because your equations became dimensionally invalid. (Always a useful check.)
    The trajectory as a whole (x from -∞ to +∞) is symmetric about the peak. What you're describing here is simply looking at different sections of it.
    If you want to maximise the range, but the target is at a different height from the start, we can do that. The equation is already set up for y to be 0 when x is zero. Set y to be the target height at [itex]\hat{x}[/itex]:
    ytgt = [itex]\hat{x}[/itex]tanθ - g[itex]\hat{x}[/itex]2/2u2cos2θ (1)
    We can now consider this to be an equation for [itex]\hat{x}[/itex] as a function of ytgt, u and θ. We want to maximise [itex]\hat{x}[/itex] wrt θ, so differentiate wrt θ and set ∂[itex]\hat{x}[/itex]/∂θ = 0:
    0 = [itex]\hat{x}[/itex]∂tanθ/∂θ - (g[itex]\hat{x}[/itex]2/2u2)∂sec2θ/∂θ
    = sec2θ - (g[itex]\hat{x}[/itex]/2u2)2sec2θ tanθ
    tan θ = u2/g[itex]\hat{x}[/itex] (2)
    Using 2 we can substitute for [itex]\hat{x}[/itex] in (1) to obtain an equation relating θ to ytgt.
     
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