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Projectile motion (the dreaded volleyball problem)

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

    A regulation volleyball court is L = 18.0 m long and a regulation volleyball net is d = 2.43 m high. A volleyball player strikes the ball a height h = 1.81 m directly above the back line, and the ball's initial velocity makes an angle theta = 48° with respect to the ground

    find:
    a.) vi the ball must be hit at to barely make it over the net
    b.)the maximum height reached by the ball in this case
    c.)vi for the ball to be hit so it directly lands on the opponent's back sideline
    d.) the maximum height for the ball in this case
    e.)maximum vi for the ball to barely make it over the net and just barely land in bounds
    (for the contact point in previous problems, what is the maximum vi)
    f.)if you hit the ball at this maximum vi, what angle should you hit it at?
    2. Relevant equations

    d=vt for the x direction

    free fall equations for the y direction

    3. The attempt at a solution

    ~i know the answer to a is 9.73 m/s what will i do with this in part b?

    ~for b. i would like to know how to set up this problem
    correct any errors for this question please

    dx=9.0m
    vx= ?
    t=9.0/vcos48

    dy= 0.62m because it it hit i.81m above the ground
    viy= ?
    a=-9.8 m/s^2

    ~part c is basically like part a but dx is 18.0m and i believe your dy is 2.43

    ~i think part d is like part b

    and i haven't tried part e or f so help will be of great appreciation! thanks to anyone who replies!
     
  2. jcsd
  3. Dec 30, 2008 #2

    CompuChip

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    Science Advisor
    Homework Helper

    What is the general formula for height (vertical distance) for a freely falling object (and watch the initial values, at t = 0).
    At what time does it reach the highest point (what happens to the vertical velocity?).
     
  4. Dec 30, 2008 #3
    The best way to do these problems is create a chart. Then from the chart, use some basic ideas and then apply the three basic equations for kinematics.

    The way you setup the chart should look like this, where o (naught) means initial. Such as xo means initial position along the x-axis.

    [​IMG]

    Things to remember: there is usually no acceleration in the x-direction, and in the y-direction will usually be gravity. Time will be the same for both.

    On a side note as well, the initial velocity will be the same for both. However, it will be a component. Since we've broke the vector up into x and y components we have V*cos(theta), and V*sin(theta) =). I will leave you to figure out which one goes to which.

    The three equations are:
    v = vo + at, x = xo + vot + (at^2)/2, and v^2 = vo^2 + 2*a*(x - xo). From here on, it's just algebra.
     
  5. Dec 30, 2008 #4
    ~for part a the question was "at which initial speed must the player hit the ball so that it just barely makes it over the net?"

    my work is as follows:

    dx= 9.0
    vx= vcos48=> .669
    t= 9/v(.669)

    dy=0.62m taking into account the height in which it was hit at
    viy= vsin48=> .743
    a=-9.8

    d=vit+1/2at^2
    .62= .669(12.11)-4.9(12.11/v)^2
    -7.48=-718.6/v^2
    v^2=96.06
    v=9.73
    my answer was correct

    ~for part b the question states "what is the maximum height above the court reached by the ball in this case?"

    my work looked like this:

    dx=9.0
    vx= 9.73(.669)=> 6.51
    t= 1.38

    dy=?
    viy= 9.73(.743)=> 7.23
    vf=0
    a=-9.8

    vf^2=vi^2+2ad
    0^2= 7.23^2+2(-9.8)(d)
    -52.27= -19.6(d)
    d= 2.66

    i put this answer into my problem set and i also added this to the height of the net and it says my answer is wrong

    ~part c asks "at what initial speed mut the ball be hit so that it lands directly on the opponent's back line?"

    i went about the problem just like part a but changed my dx to 18.0m and my dy to 2.43m giving me an answer of 14.44 and my problem set says it is wrong

    could you explain why these answers are wrong?
     
  6. Dec 31, 2008 #5

    Doc Al

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    Staff: Mentor

    Here you found the time to reach the net, which is not needed.

    This looks OK.

    Why add the height of the net? Add the initial height of the ball.

    I don't know anything about volleyball, but I assume the back sideline is at the other end of the court. Again, you used the height of the net instead of the initial height of the ball.
     
  7. Dec 31, 2008 #6
    ok so for part b my answer is now 4.47

    for part c my problem should look like this
     
  8. Dec 31, 2008 #7
    dx=18
    vx=vcos48
    t=26.9

    dy=0
    viy=vsin48
    a=-9.8
    then use d=vit+1/2at^2?
     
  9. Dec 31, 2008 #8
    i mean dy=1.81 (whoops
     
  10. Dec 31, 2008 #9

    Doc Al

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    Staff: Mentor

    You mean: t = 26.9/v

    Yes, except that dy ≠ 0.
     
  11. Dec 31, 2008 #10
    that dy = 1.81 or is it negative?
     
  12. Dec 31, 2008 #11

    Doc Al

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    Staff: Mentor

    It's negative.
     
  13. Dec 31, 2008 #12
    so here's my new work:

    d=vit+1/2at^2
    1.81= vsin48(26.9/v)-4.9(26.9/v)^2
    1.81=19.99-(3545.7/v^2)
    -18.1=-3545.7/v^2
    v^2=195.9
    v=13.99

    my problem set is saying this answer is wrong, is there a flaw in my math?
     
  14. Dec 31, 2008 #13
    wait nevermind i didn't make that negative
     
  15. Dec 31, 2008 #14
    this is my new work

    -1.81=vsin48(26.9/v)-4.9(26.9/v)^2
    -1.81=19.9-(3545.7/v^2)
    -21.8=-3545.7/v^2
     
  16. Dec 31, 2008 #15
    v^2=162.64
    v=12.75
     
  17. Dec 31, 2008 #16
    ok so for part d i have to find the maximum height the ball will reach in this case so my work will be:

    vf^2=vi^2+2ad
    0=12.75^2+2(-9.8)d
    -162.6=-19.6d
    divided by -19.6
    d=8.29
    8.29+1.81
    d=10.01
     
  18. Dec 31, 2008 #17
    ...but it is wrong, and i don't know why
     
  19. Jan 1, 2009 #18

    Doc Al

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    Staff: Mentor

    Use the vertical component of the velocity.
     
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