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Work Energy Theorem A Baseball thrown from a roof

  1. Mar 13, 2012 #1
    1. The problem statement, all variables and given/known data
    A baseball is thrown from the roof of 21.7 -tall building with an initial velocity of magnitude 13.2 and directed at an angle of 55.0 above the horizontal.

    What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance.

    What is the answer for part (A) if the initial velocity is at an angle of 55.0 below the horizontal?

    2. Relevant equations
    PE=mgh
    KE= 1/2mv^2
    KE2=K1+W


    3. The attempt at a solution

    From the prompt I gather that they are suggesting use of the Work Energy Theorem, however, every equation for KE or PE I can think of requires a mass in order to solve for the unknown variable. I've gone over the example my professor did in class, but in both of the examples he had a mass of the particle that was being thrown.

    I'm not sure how to approach this problem. A suggestion on where to start would be much appreciated. I also think that once I am on the right path I'll be able to get the remainder of the problem completed.
     
  2. jcsd
  3. Mar 13, 2012 #2
    Hi MissEuropa welcome to PF

    the work energy theorem says that the work done on an object is equal to the change in kinetic energy of the object. What work is being done on the baseball as it flies through the air? And also imagine the path of the baseball.
     
  4. Mar 13, 2012 #3
    Thanks for the welcome.

    The work being done on the baseball would be done by the person throwing it as W=F*d and the work done by gravity.

    The path of the baseball would be parabolic in shape.
     
  5. Mar 13, 2012 #4
    So the ball starts off going up, hits a peak, and then goes back down to the ground. In other words, it starts off with some kinetic energy and potential energy, hits a peak of maximum potential energy and zero kinetic energy, and then falls to the ground until it gains maximum kinetic energy and zero potential energy (relative to the ground)

    in other words, the work done by gravity on the ball as it "pulls" the ball down to the ground from its maximum height is equal to the change in the ball's kinetic energy

    as for how much mass the baseball has, if you write down the equations, I think you'll find out why it wasn't given
     
  6. Mar 13, 2012 #5
    oh and you don't need to worry about the work done by the person throwing it at first, in this situation we just say that the ball has an initial velocity

    also, is the initial velocity at 55 degrees above or below the horizontal? You have it written down both ways. Or is the question asking for both cases?
     
  7. Mar 13, 2012 #6
    I have drawn a free body diagram and written the equations that I can find, but I'm still not seeing how to approach this without the mass.

    And yup, it's asking for both cases, first 55 degrees above the horizontal, then 55 degrees below.
     
    Last edited: Mar 13, 2012
  8. Mar 13, 2012 #7
    what you want to do in this problem is use conservation of energy based on the work energy theorem.

    For the first case, the ball is going to reach a maximum height, and then it is going to fall down to the ground. What is the change in kinetic energy from the peak to the bottom?

    For the second case, the ball is going to start out with some kinetic energy and then it's gonna fall to the ground. What's the change in kinetic energy from start to finish?

    Gravity does work in both situations, and the work that gravity does is equal to the change in the kinetic energy of the ball as it goes from start to finish.


    the attached image is really the only thing you need to think about for this problem. The red trajectory is the path that the ball takes in the first case. The blue trajectory is the path that the ball takes in the second case.
     

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