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(projectile motion) calculation problem

  1. Jul 29, 2006 #1
    Question: A ball with a velocity of Vo is thrown from the bottom of an inclined plane (point O) with an angle of [tex] \phi [/tex]. The inclined plane has an angle of[tex]theta[/tex] . The ball lands on top of the inclined plane on a point A. Find the length of OA and its maximum value.

    My calculations:
    1. [tex]x = V_{0}\cos (\phi) t[/tex]
    2. y=Vo*sin[tex]\phi[/tex]t -0.5*g[tex]t^2[/tex]
    3. y=xtan[tex]\theta[/tex]

    From the above equations, I got t=[2Vo*sin[tex](\phi-\theta)[/tex]/[gcos[tex]\theta[/tex]]
    So OA= xsec[tex]\theta[/tex]

    But the answer is
    OA= [tex]Vo^2[/tex]*[sin[tex](2\phi-\theta)[/tex]-sin[tex]\theta[/tex]]/[g[tex]cos^2(\theta)[/tex]]

    And I don't know how to tell from my calculations what the max value is, and why it's different from the answer...
    Can somebody help?

    Last edited by a moderator: Apr 22, 2017
  2. jcsd
  3. Jul 29, 2006 #2


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    Change your tags to [ tex ] .... [ /tex ] (without the spaces obviously) :wink:

    Click on this equation below to see how it is coded;

    [tex]x = V_{0}\cos (\phi) t[/tex]
    Last edited: Jul 29, 2006
  4. Jul 29, 2006 #3
    thanks for teaching me latex! ^_^
  5. Jul 29, 2006 #4


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    How did you arrive at this? Would you mind posting your steps?
  6. Jul 30, 2006 #5
    from 1. and 3. ->
    y=[tex]V_{0}\cos (\phi) t[/tex]*tan[tex]\theta[/tex]
    then take the above result and plug it into 2. getting:
    y=Vo*sin[tex]\phi[/tex]t -0.5*g[tex]t^2[/tex]=[tex]V_{0}\cos (\phi) t[/tex]*tan[tex]\theta[/tex]

    Rearranging the equation and using the formula
    sin2[tex]\phi[/tex]=2sin[tex]\phi[/tex]cos[tex]\phi[/tex] to simplfy the equation makes the result for
  7. Jul 31, 2006 #6
    Up to this point, everything is fine.
    You seem to have made an error here. What is the expression for x?
    Last edited: Jul 31, 2006
  8. Jul 31, 2006 #7
    Sorry! I mistyped!!!
    It should be
    OA= Vo*cos[tex]\phi[/tex]*2Vo*sin[tex](\phi-\theta)[/tex]*sec[tex]\phi[/tex]/[gcos[tex]\phi[/tex]]
  9. Jul 31, 2006 #8
    Do you mean OA= Vo*cos[tex]\phi[/tex]*2Vo*sin[tex](\phi-\theta)[/tex]*sec[tex]\theta[/tex]/[gcos[tex]\theta[/tex]]?

    If yes, then you are on the right track. Compare your answer with the one given in the book. What is the last step you need to take?
    Last edited: Jul 31, 2006
  10. Aug 1, 2006 #9
    I think that I have to use some kind of trignometric formula to get the answer in the book?
  11. Aug 1, 2006 #10
    the ans should come out to be
    R=(2u<sq (only u)>sin<sq>(a-b)cos<sq>a)/(gcosb)<sq>

    the max value will be when a= 45-b/2

    where a = angle of projection + angle of inclination(b)
    angle of inclination=b

    if u ask i will mail u the proof
  12. Aug 1, 2006 #11
    A Better solution

    well i have written a different solution
    i am taking axis along the incline and perpendicular to the incline then taking components of g(acceleration due to gravity) along and perpendicular to the incline and then applying kinematics equations
    anywayshttp://helpjee.blogspot.com/2006/08/inclined-plane-problem_01.html" [Broken]
    :smile: :smile:
    Last edited by a moderator: May 2, 2017
  13. Aug 3, 2006 #12
    Wow! thank you for the different approach!
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