physics

A 3kg basketball is dropped from the top of a 120m building. if all of the energy is converted into KE, what is the velocity of the ball just before it hits the ground

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asked by ShaSha
  1. potential energy = m g h = 3(9.81)(120)

    Ke = .5 m v^2 = .5 (3)(v^2)
    so
    v^2 = 2 * 9.81 * 120

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    posted by Damon
  2. I am lost...please explain

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    posted by ShaSha
  3. well, yo9u can just memorize the formula. Most of us know it.
    For something dropped with no air resistance from height h
    v = sqrt (2 g h)

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    posted by Damon
  4. Let me see if I get this:

    KE=1/2(m)(v^2)

    but because I do not know the velocity, I would use the formula v=sqrt (2 g h)

    v=sqrt (2 g h)
    =sqrt (2*9.8*120)
    =sqrt (2352)

    then I would use the velocity 2352 for the formula
    KE=1/2(m)(v^2)
    =1/2 (3)(2352)
    KE= 3528

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    posted by ShaSha
  5. You could do it the way I did it:
    Potential energy lost = kinetic energy gained
    m g h = (1/2) m v^2
    g h = (1/2) v^2
    2 g h = v^2
    v = sqrt (2 g h)

    However many students and teachers have just plain memorized
    v = sqrt (2 g h)
    because it is used so frequently.

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    posted by Damon
  6. Thank you so very much

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    posted by ShaSha

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