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Projectile Motion and maximum height

  1. Mar 30, 2009 #1
    1. The problem statement, all variables and given/known data
    A projectile is launched with a speed of 40 m/s at an angle of 60 degrees above the horizontal. Use conservation of energy to find the maximum height reached by the projectile during its flight.

    2. Relevant equations
    KEi+PEi=KEf+PEf


    3. The attempt at a solution
    i am not sure where vf would come from. Would it be just Vi*cos(60)?
    I am not sure why that makes sense.
    Does Vix=Vf?...and why?
     
  2. jcsd
  3. Mar 30, 2009 #2

    LowlyPion

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    Assume your potential energy is 0 when you launch and your kinetic energy in the y direction is what? 1/2*m*Viy2?

    And at the height it has no vertical kinetic energy and the potential energy is what?

    What is the vertical component of V? (Hint: it's not Vi*cos(60))
     
  4. Mar 30, 2009 #3
    why do i need the vertical component of Vi

    I dont know how to find the V at the heighest point...

    At the heighest point is the V in vertical direction 0 anyway?
     
  5. Mar 30, 2009 #4
    I think i solved this out...i got 61m???
     
  6. Mar 30, 2009 #5

    LowlyPion

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    Yes your final V is 0.

    But your initial V is the vertical component of V, as that is the component of V that is affected by gravity ... you know, where that potential energy is building.
     
  7. Mar 30, 2009 #6
    Thank you...can you check if my answer is right?
     
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