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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]
    =Vo*cos[tex]\phi[/tex]*sec[tex]\theta[/tex]/(gcos[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?

    http://img97.imageshack.us/my.php?image=inclinedplanesu8.png
     
    Last edited: Jul 29, 2006
  2. jcsd
  3. Jul 29, 2006 #2

    Hootenanny

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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

    Hootenanny

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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
    t=[2Vo*sin[tex](\phi-\theta)[/tex]/[gcos[tex]\theta[/tex]]
     
  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
    Yes...
    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

    Hello
    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
    anyways here is the entire solution
    bye
    :smile: :smile:
     
  13. Aug 3, 2006 #12
    Wow! thank you for the different approach!
     
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