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What is the ball's maximum height?

  1. Jan 2, 2012 #1
    1. The problem statement, all variables and given/known data
    A ball is thrown upward.
    What is its maximum height? Its initial
    vertical speed is 10.8 m/s and the acceleration
    of gravity is 9.8 m/s^2
    Neglect air resistance.

    Answer in units of m


    2. Relevant equations

    dfinal=dintial+vfinalΔt+ 1/2averageaccelerationΔt^2 or v^2final=v^2initial +2averageaccelerationΔd

    3. The attempt at a solution
    14.2463
     
    Last edited: Jan 2, 2012
  2. jcsd
  3. Jan 2, 2012 #2
    No remember that since the object is travelling up it will have a negative speed therefore:

    Vf^2=u^2+2is
     
  4. Jan 2, 2012 #3
    what does u and is stand for?
     
  5. Jan 2, 2012 #4
    u and v are vectors v is the final speed which in your case will be 0 because the kinetic energy will be 0 at the maximum height so you will be left with 0=speed of the ball -2(speed of gravity)s.
     
  6. Jan 2, 2012 #5
    so how do i use this to find the height?
     
  7. Jan 2, 2012 #6
    0=(initial speed)^2-2(speed of gravity)s then once you have this down just solve the equation.
     
    Last edited: Jan 2, 2012
  8. Jan 2, 2012 #7
    I am solving for Vf^2 which is vertical speed final?
    so I plug in 0-2(10.8)=-21.6?
     
  9. Jan 2, 2012 #8

    gneill

    User Avatar

    Staff: Mentor

    Your second equation looks promising. In more compact terms:

    vf2 = vi2 + 2*a*d

    where vi is the initial velocity, vf the final velocity, a is the acceleration and d is the distance. The distance d is what you're looking for. What values will you assign to the other variables?
     
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