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A problem a bout energy

  1. May 1, 2010 #1
    when a ball with 4 kg in thrown upwards at a velocity of 40m/s, at what height is the KE equal to PE?
     
    Last edited: May 1, 2010
  2. jcsd
  3. May 1, 2010 #2
    Do you have any relevant equations yet?
    Also, I assume it is being thrown from a y(0)= 0m?
     
  4. May 1, 2010 #3
    It is like KEi+PEi=KEf+PEf and I need to find at which height they are equal

    yes it is thrown from y=0m
     
  5. May 1, 2010 #4
    [tex]\frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2 + mgh_2[/tex]
    As the equation shows, they are equal at the height that takes half of the initial kinetic energy.
     
  6. May 1, 2010 #5
    can you please solve it xcvxcvvc?
     
  7. May 1, 2010 #6
    Solve what? You don't use the entire equation I wrote. You solve the equation I defined with my words.
     
  8. May 1, 2010 #7
    Have you tried anything yet?
     
  9. May 1, 2010 #8
    yes it should be h=40 m to be equal.
    Because the KEi=3200 J and it should take the half of it to be equal,which means 1600J
     
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