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How to determine X and Y coordinates of launched ball?

  1. Jul 18, 2007 #1
    1. The problem statement, all variables and given/known data
    Lets say one were to launch a ball at 60 degrees at 80m/s. How would one determind the x and y coordinates at time intervals of the ball?

    2. Relevant equations
    I'm not sure of equations that should be used.

    3. The attempt at a solution
  2. jcsd
  3. Jul 18, 2007 #2


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    1) F= ma

    2) Force and acceleration are vectors: you can look at X (horizontal) and Y (vertical) separately.

    Neglecting air resistance, the only force on the ball after it has been "launched" is gravity, -mg, vertically.

    From this point, I can think of several different ways to procede but, since you have shown no work at all, I don't know which would be appropriate! If you have not taken calculus, then there are probably specific equations given in your textbook.
  4. Jul 18, 2007 #3
    I'd like to use the equation:

    Δx = Vavg x Δt
    Δy = Voy Δt + ½ g Δt^2

    The thing is, when I use these equations the values just keep getting bigger and bigger....I don't know what/how to manipulate the equation to get a negative parabola.
  5. Jul 18, 2007 #4
    Relevant equations:

    [tex]x = x_i + v_i t + \frac{1}{2} a t^2[/tex]
    [tex]y = y_i + v_i t + \frac{1}{2} a t^2[/tex]

    *NOTE: the velocity and acceleration must be the vector portions:

    [tex]\Delta s[/tex] = total displacement
    [tex]s = \sqrt{x^2 + y^2}[/tex]
    [tex]x = s cos \theta[/tex]
    [tex]y = s sin \theta[/tex]

    However, velocity is a vector, like s, your displacement. So, you apply trig, record the correct velocity in the y-direction and in the x-direction.
    Last edited: Jul 18, 2007
  6. Jul 18, 2007 #5
    Thanks Pheonix that helps a bunch. I'm having the same problem as before. I'm sure it's a simple mis understanding, but since gravity is the sole effect on the ball, the velocity obviously has to decrease towards the peak. Using the equations above, as time increases, so does velocity. How do I have the velocity values go back down as if I were to actually do the lab?
  7. Jul 18, 2007 #6
    You are on the right track. But I suggest you first consider the following problem: You throw a ball vertically up in the air with the initial velocity Voy.
    How does the equation look like that describes the motion of the ball?
  8. Jul 18, 2007 #7
    I don't know....new to this whole Physics thing.
  9. Jul 18, 2007 #8
    Use the equation you gave for Δy.
    Suppose you throw a ball vertically up in the air with 80 m/s.
    How does Δy look like?
  10. Jul 18, 2007 #9


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    Are you putting g=-9.8m/sec^2? Emphasis on the NEGATIVE.
  11. Jul 23, 2007 #10
    agreeing with goldpheonix, the ket is to seterate the initial velocities into components with trig. & depending on question resolving the final velocity at the given time peroid with pythagoras

    havn't done projectiles in a while though.
  12. Jul 24, 2007 #11
    The velocity and acceleration must oppose each other as dick said
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