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Projectile Motion - Volleyball

  1. Sep 29, 2008 #1
    1. The problem statement, all variables and given/known data

    A regulation volleyball court is L = 18.0 m long and the net is d = 2.43 m high. A volleyball player strikes the ball a height h = 1.61 m directly above the back line, and the ball's initial velocity makes an angle θ = 49° with respect to the ground. At what initial speed must the ball be hit so that it just barely makes it over the net? (Assume the volleyball is a point object and is hit so that its path is parallel to the sideline, as seen from directly above the court.)

    3. The attempt at a solution

    I am really lost at how to figure out this one without time. I tried using the following equation since you are given the height it has to reach, and the acceleration (-9.8) and I think the final velocity would be zero:
    [tex]
    v^2 = v_0^2 + 2 a \Delta x
    [tex]

    but it doesn`t seem to work. I am thinking I have to calculate the time somehow, but I really can`t figure out how to go about doing that. Can anyone help?
     
  2. jcsd
  3. Sep 29, 2008 #2

    LowlyPion

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    Homework Helper

    It's not quite so simple.

    First you need to determine the Vertical and Horizontal components of velocity based on the 49 degrees.

    Then you can develop the x and y motion equations using the 18 m as the range and the height of the net as the height at the instant of clearance.

    The equations

    [tex] Y_{net} = Y_{serve} +V_y*t -1/2*g*t^2[/tex]

    [tex] X_{to.net} = V_x*t [/tex]
     
  4. Sep 29, 2008 #3
    Ok, how though, do I find the time? Or should I solve for t in the x equation and sub it in?
     
  5. Sep 29, 2008 #4

    LowlyPion

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    You should have equations with only initial velocity and time unknown.

    Eliminate 1 and solve for the other.
     
  6. Sep 29, 2008 #5
    Ok so I tried that, and I think I have an error somewhere because it doesn't work. Here is what I did:

    t = 9/(Vsin49)

    2.43 = 1.61 + Vcos49t - 0.5*9.8t^2
    0.82 = Vcos49 * 9/Vsin49 - 4.9 * (9/Vsin49)^2

    V = 7.9 m/s

    I can't seem to find what I did wrong though..
     
  7. Sep 29, 2008 #6

    LowlyPion

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    I calculate your equation differently.

    Using Sin49 = .755 and Cos49=.656
    I get: t=11.92/v

    Substituting:
    .82 = .656(11.92) - 4.9(11.92/V)2
    .82 = 7.82 - 4.9(11.92/V)2
    7 = 4.9(11.92/V)2
    V2 = 4.9(11.92)2/7 = 99.46
    V = 9.97 m/s
     
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