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Projectile Motion Up An Incline Plane

  1. Jun 14, 2009 #1
    1. The problem statement, all variables and given/known data
    A projectile is fired up an incline (incline angle φ) with an initial speed vi at an angle θi with respect to the horizontal (θi > φ). (a.) Show that the projectile travels a distance d up the incline, where

    d = 2*vi^2*cosθi*sin(θi-φ) / g*cos(φ)^2

    (b.) For what value of θi is d a maximum, and what is the maximum value?


    2. Relevant equations



    3. The attempt at a solution
    Ok I am back and unbelievably got part a, but I have no idea how to do part b. I know 45 is the way to get maximum range, but I don't think that is the right angle, because we want to maximize x^2+y^2. Any help would be appreciated!
     
  2. jcsd
  3. Jun 14, 2009 #2
    Here's a picture.
     

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  4. Jun 14, 2009 #3

    ideasrule

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    Homework Helper

    You have an equation relating d, which you want to maximize, with θi. It is:
    d = 2*vi^2*cosθi*sin(θi-φ) / g*cos(φ)^2

    Do you know calculus? If you do, just derive d with respect to θi, set the derivative to zero, and solve for θi. It shouldn't be too hard if you remember a few trig identities, like sin(θi-φ)=sin(θi)cos(φ)-sin(φ)cos(θi).

    If you don't know calculus, try to use trig identities to rearrange the formula for d so that it's easy to identify the maximum. Remember that vi and φ are constants, so the only part you need to maximize is cosθi*sin(θi-φ). This is much harder mathematically than using calculus; I can't immediately see how you would go about manipulating cosθi*sin(θi-φ).
     
  5. Jun 15, 2009 #4
    Ah this derivative is too hard for me to get. I'll try your second way. Thank you.
     
  6. Jun 15, 2009 #5

    ideasrule

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    The calculus way is actually very simple:

    Derive cosθi*sin(θi-φ), because everything else contains only constants. Use the product rule and you'll get cosθi*cos(θi-φ)-sinθi*sin(θi-φ). Recall that cos(a+b)=cos(a)cos(b)-sin(a)sin(b), so:

    cosθi*cos(θi-φ)-sinθi*sin(θi-φ)=cos(2θi-φ)=0

    What must 2θi-φ be in order for cos(2θi-φ) to be zero?

    Without using calculus:

    cosθi*sin(θi-φ)=cosθi*(sinθi*cosφ-sinφcosθi)
    =cosθi*sinθi*cosφ-sinφ*cos^2 θi
    =cosθi*sinθi*cosφ-sinφ(cos(2θi) +1)/2, since cos(2θi)=2cos^2 (θi) - 1
    =1/2 sin 2θicosφ - 1/2 sinφcos(2θi) - 1/2 sinφ
    =1/2 (sin 2θicosφ-sinφcos(2θi)-sinφ)
    The factor 1/2 can be neglected because it's a constant; so can sinφ for the same reason. Our job now is to maximize sin 2θicosφ-sinφcos(2θi), which can be rewritten as
    sin (2θi-φ). The maximum possible value of a sine function is 1, so sin (2θi-φ)=1. What must 2θi-φ be?
    Make sure that your answers for the calculus and non-calculus methods agree!
     
  7. Jun 15, 2009 #6
    Ok I got θi=45+φ/2. Thank you.
     
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