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Kinetic Energy of diver

  1. Oct 4, 2004 #1
    Kinetic Energy, Help!!!

    A 66.0 kg diver is 4.90 m above the water, falling at speed of 8.20 m/s. Calculate her kinetic energy as she hits the water. (Neglect air friction)

    Then the problem hints:Use conservation of mechanical energy.

    So i'm thinking that conservation is K_f+U_f=K_i+U_i but there must something wrong with my calculations>
    K=1/2(m*v^2) and U=mgh right?
    Last edited: Oct 4, 2004
  2. jcsd
  3. Oct 4, 2004 #2
    ok let's say that delta (the triangle) is represented by d in my calcultions

    dK + dU = 0 conservation of energy

    KE (final) - KE (Initial) + PE (Initial) - PE (final) = 0

    sinceu want a value for KE final

    PE (final) = m g (0) = 0 becuase as diver hits water height is zero

    KE final = PE (initial) + KE (initial)

    now sub in and see waht you get
  4. Oct 4, 2004 #3
    From KE_final=PE(initial)+KE(initial) i get:
    (66.0)(9.8)(4.90)+1/2(66.0)(8.20^2)= 5388.2 and it was right thanxs a lot
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