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Projectile's launch speed?

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data

    The highest barrier that a projectile can clear is 15.0 m, when the projectile is launched at an angle of 13.0° above the horizontal. What is the projectile's launch speed?
    2. Relevant equations
    kinematics equations


    3. The attempt at a solution

    delta y=15m v0x= vcos13
    ay=-9.8m/s^2 ax=0m/s^2

    im not sure how to use the information that i have to solve the problem
     
  2. jcsd
  3. Mar 2, 2009 #2

    LowlyPion

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    What does clearing a maximum height barrier suggest?

    Will that be the highest point of its trajectory?
     
  4. Mar 2, 2009 #3
    yes 15m is the highest point i just don't know how to use that to solve the problem
     
  5. Mar 2, 2009 #4

    LowlyPion

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    What will the vertical component of velocity need to be if it is only to go 15 m high?
     
  6. Mar 2, 2009 #5
    vsin13?
     
  7. Mar 2, 2009 #6

    LowlyPion

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    Yes it will equal that.

    But they ask you to find Vo to begin with.
     
  8. Mar 2, 2009 #7
    i don't know how to find Vo with the given info.. thats my problem
     
  9. Mar 2, 2009 #8

    LowlyPion

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  10. Mar 2, 2009 #9
    so use v^2=Vo^2+2aX?? But i still dont know what v or Vo are or how to get them...
     
  11. Mar 2, 2009 #10

    LowlyPion

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    So at max height, what will the vertical velocity be again?
     
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