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Velocity as Function of Time

  1. Feb 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Okay so, a rock is throw from an initial height, ho, at a speed, vo, and an angle [theta] from the horizontal. Find the rocks velocity as a function of time.

    2. Relevant equations

    Vxo=(Vo)(cos[theta])
    Vyo=(Vo)(sin[theta])
    Acceleration is constant.

    3. The attempt at a solution

    Well I know that V(t)=Vo+at

    But since the rock is thrown from an angle I got confused. Do I have to replace the Vo with the two equations for the x and y component of velocity? Since I can solve for Vo, but then it is equal to two different things.
    Thanks!
     
  2. jcsd
  3. Feb 18, 2009 #2
    Find the vertical component of the velocity and find the horizontal component of the velocity.

    With the vertical component you can find out how long it takes for the ball to reach maximum height, where vf = 0

    Then find out how high it rises and add the initial height from which it was thrown from and you will be able to get the total flight time.
     
  4. Feb 18, 2009 #3
    How do I find out the components if I am not given any numbers?
     
  5. Feb 18, 2009 #4

    sfh

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    You are not going to have a number as your answer. It sounds as if you will have an algebraic expression that includes the variable t for time.

    You have the right start, certainly find the components in the x- and y- directions. Then treat them like vector and add them together. Note that your "acceleration is constant" statement is incomplete. You have a lot more information than that.
     
  6. Feb 18, 2009 #5
    So is my final equation?:

    v(t)=[(Vxo/cos[theta])+(Vyo/sin[theta])]+at
     
  7. Feb 18, 2009 #6

    sfh

    User Avatar

    No. You'll need to draw a picture with the x- and y- component vectors. To find the magnitude, you'll need the pythagorean theorem. To find the angle (also a function of time), you'll need a trig identity.
     
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