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Problem regarding projectile motion on a inclined plane

  1. Dec 30, 2015 #1
    1. The problem statement, all variables and given/known data
    A projectile is thrown on an inclination plane with incline angle phi with an inital velocity V0 on an angle theta. Find the range of the projectile; treat the hypotenuse of the inclined plane as the x-axis. For what value of initial velocity angle is the range maximum, and what is that range.
    upload_2015-12-30_19-21-3.png

    2. Relevant equations
    All 2D kinematics equations.

    3. The attempt at a solution
    Firstly I used the equation y(t) and expressed flight time as 2Vy/-gy (g is negative itself, the minus in front of it makes it positive), from there I wrote t=(2V0*sintheta)/(-g*cosphi). Then I plugged that into the equation for range which is D=Vx*t+1/2gx*t^2 where gx is the x-axis component of g. By simplifying the expression I obtained D=(2V0^2*sintheta*(-costheta*cosphi+sintheta*sinphi))/(g*(cosphi)^2). Now everything is correct except the bolded part which should be costheta*sin(theta-phi). Help please.
     
  2. jcsd
  3. Dec 30, 2015 #2

    TSny

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    Note that the angle θ in the figure is measured from the horizontal, not from the inclined surface. It appears that you used Vy = V0 sinθ. Is this correct?

    (If you click on the Σ symbol on the formatting tool bar, you can access more math symbols, like θ.)
     
  4. Dec 30, 2015 #3
    Yes! Sorry for the careless mistake, I found another attempt using the correct equation for t, but still everything is correct except the trigonometric part which then equals to sin(θ-Φ)*(sin(θ-Φ)*sinθ-cos(θ-Φ)*cosΦ), and I used Wolfram Alpha to check if this was equal to the correct cosθ*sin(θ-Φ) and it was not.
     
  5. Dec 30, 2015 #4

    TSny

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    You have sin(θ-Φ)*(sin(θ-Φ)*sinθ-cos(θ-Φ)*cosΦ). I suspect that the mistake is the angle θ in red. Check that.
     
  6. Dec 30, 2015 #5
    Yes! Thank you! It should be sinΦ. Thanks!
     
  7. Dec 30, 2015 #6
    Could you please prove algebraically that sin(θ-Φ)*(sin(θ-Φ)*sinΦ-cos(θ-Φ)*cosΦ)=cosθ*sin(θ-Φ), and also if you could give me some hints for finding the angle for which that expression is going to be maximum. Thank you.
     
  8. Dec 30, 2015 #7

    TSny

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    Let α = θ-Φ and β = Φ. Write sin(θ-Φ)*sinΦ-cos(θ-Φ)*cosΦ in terms of α and β. Then see if it reminds you of a well-known trig identity.

    Write D in the form D = C⋅f(θ), where C is independent of θ. D is maximized when f(θ) is maximized
     
  9. Dec 30, 2015 #8
    Thanks for the hint, I proved it. If you could just tell me what would you do to find the angle for maximum f(θ) because my brain is really blocking currently. And if you can, do not use derivatives.
     
  10. Dec 30, 2015 #9

    TSny

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    Try using another trig identity: http://www.sosmath.com/trig/prodform/prodform.html
     
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