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Kinetic Energy of baseball

  1. Jun 17, 2007 #1
    1. The problem statement, all variables and given/known data
    an outfielder throws a 2.15 kg baseball at a speed of 113 m/s and an initial angle of 11.2 degrees. What is the kinetic energy of the ball at the highest point of its motion.


    2. Relevant equations
    KE = 1/2mv^2

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


    3. The attempt at a solution

    KE = 1/2 (2.15 kg) (113* sin 11.2)^2
    KE =517.866

    My method seems too simple. Should I be using the conservation of energy equation and if so, how should I go about doing it?
     
  2. jcsd
  3. Jun 17, 2007 #2
    Almost right - have a look at you sine/cosine.

    You know that you can consider the velocity components in the horizontal and vertical directions independently. What are the forces acting along each direction of the velocity components? When the ball is at its highest point, what can you say about the ball's vertical velocity? And what about the horizontal velocity?

    As I say, you've almost got it right, just look at splitting the velocity into components again. :smile:
     
  4. Jun 17, 2007 #3
    when the ball is at its highest point, the ball's vertical velocity would be zero? Should I have used: KE = 1/2 (2.15 kg) (113* cos 11.2)^2 instead?
     
  5. Jun 17, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    That is correct.
     
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