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Application of quadratic functions to volleyball

  1. Jan 2, 2015 #1
    1. The problem statement, all variables and given/known data
    A player hits a volleyball when it is 4 ft above the ground with an initial vertical velocity of 20 ft/s (equation would be h = -16t2 + 20t + 4). What is the maximum height of the ball?

    2. Relevant equations
    quadratic formula

    3. The attempt at a solution
    t = -20 ±√202 - 4(-16)(4) / 2(-16)
    t = -0.175390 and 1.425390

    I don't think my answer is correct, but I don't know what else to do. And I don't know how to find the maximum height of the ball using the answer above :( Any help would be greatly appreciated, thank you!!
     
  2. jcsd
  3. Jan 2, 2015 #2

    haruspex

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    In the most usual formulation, there are five variables in the SUVAT equations (the equations that apply for constant acceleration). There are five equations, each involving four of the variables.
    Typically, you know three values and want to find a fourth. Select the SUVAT equation which involves those four variables.
    Which four variables are of interest here?
    Which SUVAT equation uses those four?
     
  4. Jan 2, 2015 #3

    lurflurf

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    rewrite the equation in vertex form

    h = -16t2 + 20t + 4=41/4-16(t-5/8)^2

    for what t does h achieve its maximum? What is the maximum?
     
  5. Jan 2, 2015 #4
    is the maximum the vertex, which in this case is (-5/8, 41/4)?
     
  6. Jan 2, 2015 #5

    haruspex

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    Wouldn't it be simpler to use a more appropriate SUVAT equation instead?
     
  7. Jan 3, 2015 #6
    I tried using a SUVAT equation :( s = ut + (1/2)(a)(t^2) but ahh it confused me :(
     
  8. Jan 3, 2015 #7

    lurflurf

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    yes, watch the minus sign, the vertex is (5/8,41/4)
     
  9. Jan 3, 2015 #8

    lurflurf

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    I just found out what that is. Very confusing. I am not sure who SUVAT is for, it confusing, hard to remember and limited in use.
    We need to try to remember 5 useless equations to see which one does not contain time
    $$h_{\mathrm{final}}=h_{\mathrm{initial}}+\dfrac{v_{\mathrm{final}}^2-v_{\mathrm{initial}}^2}{2a}$$
    or
    $$S=S_{\mathrm{initial}}+\dfrac{V^2-U^2}{2A}$$
    is the winner! Then we can figure out what each variable should be.
     
  10. Jan 3, 2015 #9
    That is the same SUVAT equation of :

    $${v_{final}}^2={v_{initial}}^2-2g\triangle y$$

    I think they derive those equations just to make the motion physics easier to calculate..,,
     
  11. Jan 3, 2015 #10

    SteamKing

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    If you have an equation which relates the height of the ball as a function of time, and you want to determine the time at which the height is a maximum, you don't solve the quadratic equation. Instead, you want to determine the time tmax when dh/dt = 0.

    With h = -16t2 + 20t + 4, what is dh/dt?
     
  12. Jan 3, 2015 #11

    lurflurf

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    ^I don't know the starters level, but this is a common type of problem for twelve year old students who do not know about SUVAT of derivatives. Neither are need to solve the problem.
     
  13. Jan 3, 2015 #12

    haruspex

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    Solving the problem requires the use of SUVAT equations, whether you know them by that name or not.
    It helps if you can remember the set of 5, but it's not essential. The one that does not involve time is equivalent to conservation of mechanical energy.
    In the present problem, you can use s = ut + (1/2)at2) but you need to determine t first. This can be done directly from the definition of acceleration, a = △v/△t; this is the same as another SUVAT equation.
     
  14. Jan 4, 2015 #13

    ehild

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    The equation h = -16t2 + 20t + 4 gives the height as function of the time. If you solve the quadratic equation -16t2 + 20t + 4=0 you get the time when the ball reaches the ground. It is the time after the ball was kicked up, so the negative time has no meaning.
    But you do not need the time, you need the maximum height. If you plot the function, you get a parabola. It has the standard form h= -16(t-a)2+b, and the maximum height is b.
    Find that form of the function by the method "completing the square" . You certainly have learnt it.
     
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