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Verticle Jump

  1. Dec 12, 2008 #1
    1. The problem statement, all variables and given/known data
    In order for a volleyball player to jump vertically upward a distance of 0.8 meters, his initial velocity must be?


    2. Relevant equations
    s=v0t + .5at2

    vf= v0 + at


    3. The attempt at a solution



    I missed the question the other day on a practice and it has been forever since I messed with this material. Those equations were given as well as the answer, the only problem is I do not know what the letters represent or even how to solve for anything with only information of 0.8 meters.

    Could someone help me learn this and what the expressions mean so I can solve it?
     
    Last edited: Dec 12, 2008
  2. jcsd
  3. Dec 12, 2008 #2
    Keep in mind the direction of g in reference to the player's direction of motion, i.e. positive or negative?

    Also, do you have another variable besides v0 that you need but don't have? Try to solve it in terms of another variable, which kills two birds with one stone.

    EDIT: Oh, and the variables mean

    s, the displacement
    v0, the initial velocity
    t, the change in time
    a, the acceleration of the body (g = 9.8 m/s^2)
    vf, the final velocity

    Always find out what the key variables mean in a lesson before solving the problems.
     
  4. Dec 12, 2008 #3
    ooo thank you so much for laying out what the variables mean...let me see what I can do now...


    Let me try and have a crack at this again.
     
  5. Dec 12, 2008 #4
    Displacement is the same as distance in this case. Your distance, vertical, is given to be 0.8 m.

    g IS your acceleration a.

    EDIT: Oh, and I saw your other thread. You must be lacking a few concepts without an instructor or textbook... When a body reaches a maximum height (i.e., it falls down instantly afterwards) in vertical motion, the velocity (final) at that instant is equal to zero.
     
  6. Dec 12, 2008 #5
    Ya...I have that down...but I am having a problem...lets see...

    vf= v0 + at
    so
    0 = v0+9.8t

    or am I doing gravity wrong?

    Cause I solved for v0 and moved over to the other equation...

    0.8 = -9.8t2 + .5(9.8)(t2)
    .8 = -9.8t2 + 4.9t2
    (Now if I already have not made an error with a before this is where I am getting stuck)

    .8 = -4.9t2

    I cannot remember how to get t alone from this point (bad I know)
     
  7. Dec 12, 2008 #6
    As your velocity approaches zero from a nonzero number, does it decrease or increase? This change in velocity is otherwise known as acceleration. Rethink putting +9.8.
     
  8. Dec 12, 2008 #7
    Ok so it is

    0.8 = 9.8t2+(.5)(-9.8)(t2
    0.8 = 4.9t2
    .163 = t2
    .4 = t


    0 = v0 - 9.8(.4)
    0 = v 0 - 3.92
    3.92 = v0


    Now all I need is more than this one problem to practice lol...you are awesome, thank you, thank you, thank you.
     
  9. Dec 12, 2008 #8
    Students helping students :D

    By the way, you could have done the problem in 2 steps instead of 3 steps if you had solved for t instead of solving for v0 first. Generally with algebra, to get rid of a variable you solve for that variable.
     
  10. Dec 12, 2008 #9
    Thank you...lol...good thing about me getting stuck on this equation...

    A) Now I know what these variables mean...
    B) Memorized some formulas lol

    I will try making up some problems later and solve in 2 steps...

    It has been sooo long since I messed with math or anything.
     
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