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High Jump

  1. Jun 3, 2007 #1
    In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole. With what minimum speed must the athlete leave the ground in order to lift his center of mass 1.55 m and cross the bar with a speed of 0.70 m/s

    I have tried solving this problem by using the formulas for kinetic and potential energy. I thought that the initial kinetic energy would equal the potential energy plus the kinetic energy in the air.

    KE = PE + KE'
    .5(m)(v^2) = m(g)(h) + .5(m)(v^2)
    Then i figured i could cancel out mass since it is in all parts
    .5(v^2) = (9.81)(1.55) + .5(.7^2)
    v= Sqrt (15.205 + .245)
    v = 30.9

    This is the wrong answer, and i am not sure what i am doing wrong. Please help.
  2. jcsd
  3. Jun 3, 2007 #2


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    Did you actually take the square root? It doesn't look like you did.
  4. Jun 3, 2007 #3
    Thank you. that was the problem! Appreciate it!
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