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Easy kinematics

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Girl kicks ball straight up, makes contact with ball 1 m above ground, initial velocity = 3 m/s
    a.) How high does the ball go?
    b.) How long until it hits the ground?
    c.) If she had given it an initial x-component of velocity of 2 m/s (same y-component), how far would the ball land from her?
    d.) With what velocity would it hit the ground?

    2. Relevant equations
    vf2 = vi2 + 2ay(yf-yi)
    vf = vi + at
    xf = xi + vit + 1/2axt2

    3. The attempt at a solution
    a.) vf2 = vi2 + 2ay(yf-yi)
    0 = 32 + 2(9.8)(yf - 1)
    -9 = 19.6yf - 19.6
    10.6/19.6 = yf
    yf = 0.5408 m
    This seems like a really small number; shouldn't it be higher?

    b.) vf = vi + at
    0 = vi + -gt
    t = vi/g
    t = 3/9.8 = 0.306 s
    0.306 s * 2 = 0.612 s (because the flight curve is symmetric)

    I'm not really sure what equations to use for parts c and d, neither am I sure if parts a and b are correct.
    Thank you very much!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 7, 2009 #2
    a) check your convention is gravity positive or negative?

    b) you need to think about where she initially kicked it from and whether time up == time down for entire trip or just for the symmetric part.

    c) What do you need to know about the ball to determine how far it traveled given no air resistance?
    hint: it's none of those equations, it's simpler than that.
     
  4. Oct 9, 2009 #3
    Thank you for your help so far.

    a.) Changing a to -9.8 gives a final y-position of 2.837 m.
    b.) Using the eqn. yf = yi + vit + 1/2at2 makes more sense because it takes into account the non-zero starting position of 1 m.
    Therefore, t = 1.174 s
    question:
    Do I have to multiply this answer by 2? (I don't think so, because doesn't the quadratic equation represent the entire parabola of the flight curve?)

    question:
    c.) I don't know what you mean by not having to use any eqn.
    I used xf = xi + vit + 1/2at2
    xf = 1 + 2(1.174) + 0 (since no a in x-direction)
    xf = 3.348 m
    Is this not correct?

    question:
    d.) Would vf = vi because there no acceleration in x-direction?
    Would the answer be 2 m/s?
     
  5. Oct 9, 2009 #4
    for c I was thinking of even more simply: d = v*t, which is just a simplified version of the one you used.

    d) you still need to take y velocity into account here...
     
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