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Projectile motion of a ball launch

  1. Oct 26, 2008 #1
    1. The problem statement, all variables and given/known data
    On earth, a ball is launched over level ground. It's vertical equation of motion is
    y(t) = -4.9(t2 - 8t - 20)

    A) Find tymax and ymax
    B) Find vyinitial and vyfinal

    2. Relevant equations
    y(t) = Yi + via + 1/2at2
    Vf = vi + at

    3. The attempt at a solution
    I tried to complete this problem by completing the square. I'm not sure I completed the square correctly.

    For the completed square I got:
    -4.9(t-4)2 + 58.4
    making tymax = 4s
    and ymax = 58.4 m

    To get tymax I set the equation equal to zero and factored
    I got
    0 = -4.9(t + 2) (t -10)
    making ti = -2 s and
    tf = 10 s

    For Vyi I took the derivative of the y(t) equation and plugged in ti
    Vyi = -9.8(-2) + 39.2
    Vyi = 58.6 m/s

    For Vyf I used the derivative and plugged in tf
    Vyf = -9.8(10) + 39.2
    Vyf = -58.8 m/s

    If I did not complete the square in the first part, please show me how to do it rather than telling me a different way to go about the problem. Thank you!!!
  2. jcsd
  3. Oct 26, 2008 #2


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    You must differentiate first and then set the result equal to zero. The gradient is obviously zero at maxima and minima.
  4. Oct 26, 2008 #3
    I must differentiate before getting the t value?
    Because I did differentiate before I got the Vi and Vf values.
  5. Oct 26, 2008 #4


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    Like I say, to find a maximum or minimum you need to find where the gradient is zero, therefore you must differentiate.
  6. Oct 27, 2008 #5
    Can you explain that a little further? Because I have no idea what you're talking about.
  7. Oct 27, 2008 #6


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    The derivative of a function tells us something about how that function behaves at certain points. At stationary points like maximums and minimums the derivative is zero. Therefore to find the point where this function is maximum you must take its derivative, set it equal to zero and solve.
  8. Oct 27, 2008 #7
    You have to find the derivative of the function and set it to equall to zero.when you set it to equall to zero your final solution will be two numbers so if the graph goes from negative to positive you will have minimum and from positive to negative you will have maximum point.but If you graph it, it will be easy.

    And one more derivative is a slope.
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