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Projectile Motion from a height

  1. Dec 3, 2008 #1
    A projectile is shot at a 35 degree angle from a height of 900 m. The it lands 2500 m away vertically. What was the initial velocity of the launch?



    I do not know what relevant equations to use- I've tried using the integrals of position vectors like Vox(t) = 2500 m and Voz(t) - 1/2 gt^2 = 900 m, but I don't know how to solve this without knowing the time of flight.
     
  2. jcsd
  3. Dec 3, 2008 #2

    G01

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    You can do this without calculus.

    HINT:

    Use the kinematic equations and set up equations the motion in both the x and y directions. This should give you enough equations to eliminate time from the problem and solve for the initial velocity.
     
  4. Dec 3, 2008 #3
    Do I use the kinematic equations Vo= at and x= .5Vo(t)?
     
  5. Dec 3, 2008 #4
    Since I'm given an angle, I feel like I should be using trig. Is there some way I can involve trig in the kinematic equations?
     
  6. Dec 3, 2008 #5
    OK, I have figured out what I need to do, but need help with the math!

    I need to solve for t in the 2 eqs: 2500 = Vocos35t and 900 = (Vosin35 - 4.9t)(t) and then set the equations equal to solve for Vo. But in the second equation, how do I solve for t?!
     
  7. Dec 4, 2008 #6

    G01

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    In the second equation you have a quadratic (Bring all terms to one side of the equation). You need to use the quadratic formula to solve for t.
     
    Last edited: Dec 4, 2008
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