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Desperate for help!

  1. May 28, 2007 #1
    a .565kg football is kicked from the ground at a speed of 22 m/s. what is its speed just as it passes between the uprights 5.0 m above the ground?

    please help!

    G:
    m= .565 kg
    v= 22m/s
    5.0 m

    r:
    velocity

    a:no kinematics's equations
     
  2. jcsd
  3. May 28, 2007 #2

    hage567

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    What about conservation of energy? What do you know about that?
     
  4. May 28, 2007 #3
    the only formulas i have are these

    Energy Total = Energy kenetic + Energy Gravity

    Ek = 1/2 mv^2

    Eg = mgh

    P=E+W

    W= mgh
     
  5. May 28, 2007 #4

    hage567

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    OK, the first of those are useful here. What do you think you should do with them? What does conservation of energy mean?
     
  6. May 28, 2007 #5
    conservation:
    energy cant be made or destroyed
    total energy of any closed system remains same
    energy can change from one from to another but total energy is same
     
  7. May 28, 2007 #6

    hage567

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    So if the ball is on the ground, what does that tell you about the gravitational potential energy? Can you find the kinetic energy the ball has right after it is kicked?
     
  8. May 28, 2007 #7
    well
    Et= ek + eg
    i know my v1 i think is 22m/s or is that my v2?

    my h is 5.0
    umm and my mass is .565

    if i understood the velocity part that would help i think
    cuz ek = 1.2 mv^2 but
    v is v2-v1 and v2 is either 0 or 22
     
  9. May 28, 2007 #8
    and no gravitational energy
     
  10. May 28, 2007 #9

    hage567

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    Your initial velocity (v1) is 22 m/s

    Think of it in two parts. The first part is when the ball in on the ground right after being kicked. The second is when the ball is at the 5 m height. Write out the energy terms for each part.

    So for first part Et1 = ek1 + eg1
    for the second Et2 = ek2 + eg2
     
  11. May 28, 2007 #10

    hage567

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    For the first part, when the ball is at ground level, that is correct.
     
  12. May 28, 2007 #11
    ET1 = 0
    Et2 = 136.73 (1/2 * .565kg * 22m/s^2)
    then...
     
  13. May 28, 2007 #12
    wait hold on i think i missed it so at 5.0 m above ground its 22m/s right
     
  14. May 28, 2007 #13

    hage567

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    Et1 is the sum of the two kinds of energy. When the ball is kicked, it has kinetic energy, and no potential energy since it is on the ground. So Et1 should be what you calculated for Et2.

    Et2 will also be the sum of the kinetic and potential energies, but this time, the potential energy is NOT zero, since the ball is now 5 m above the ground. So you are given enough information to find the potential energy of the ball at this point, and you need to find the new kinetic energy.
     
  15. May 28, 2007 #14

    hage567

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    NO. At 5 m above the ground, you don't know the velocity. That is what you are trying to find.
     
  16. May 28, 2007 #15
    k i calculate Eg to be 27.685
    then i looked at kinetic energy...but that leaves 2 variables velocity and kinetic energy in the formula

    Ek = 1/2 mv^2

    hmm...
     
  17. May 28, 2007 #16

    hage567

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    You know that the initial kinetic energy of the ball must be equal to the sum of the kinetic and potential energy when the ball goes through the uprights. So you can find the final kinetic energy of the ball, since you know the potential energy. Once you have that, you can solve for the velocity. There is only one unknown here.
     
  18. May 28, 2007 #17
    sorry i dont understand... how does Ek1 = Ek + Eg and even then i dont no Ek:confused:
     
  19. May 28, 2007 #18

    hage567

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    Because energy must be conserved. If the ball ONLY has kinetic energy to begin with, that means when it is above the ground at 5 m, some of that initial kinetic energy goes into gravitational potential energy, and some remains as kinetic energy. You are trying to find the FINAL kinetic energy, it will be different that the initial kinetic energy. Once you find the final kinetic energy, you can then find the velocity the ball has at 5 m above the ground.
     
  20. May 28, 2007 #19
    so therefore Ek2 =
    et1-eg
    Et1 = 136.73 (1/2 * .565kg * 22m/s^2)

    so ek2 =109.045 (136.73-27.685)
    and so there fore......
     
  21. May 28, 2007 #20
    i got 19.6468827 as velocity
     
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