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Homework Help: Help with 1 problem

  1. Oct 10, 2008 #1
    1. An outfield throws a 1.51 kg baseball at a speed of 118 m/s and an initial angle of 41.2 degrees. What is the kinetic energy of the ball at the highest point of its motion? Answer in units of J.

    2. Relevant equations:None

    3. The attempt at a solution: I used the equation K=1/2mv^2
    1/2(1.51)(118)^2= 10,512.62 J but answer was wrong. Any ideas what I did wrong?
  2. jcsd
  3. Oct 10, 2008 #2


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    You calculated the initial kinetic energy. They are asking for the kinetic energy when the ball is at the height of it's trajectory. v will be different than 118 m/s there.
  4. Oct 10, 2008 #3
    So how would I find V, as the final velocity?
  5. Oct 10, 2008 #4
    Well at it's highest point in it's trajectory, it's not rising nor falling, am I right?

    Therefore the y component of velocity = 0.

    Take it from there.
  6. Oct 11, 2008 #5
    sorry to inform you but the velocity isn't zero.
  7. Oct 11, 2008 #6


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    Rake-MC is correct that the y component of velocity is zero at the highest point.

    redhot209 is also correct, the velocity is not zero.

    Both statements are correct, since velocity and y-component of velocity are not the same thing.
  8. Oct 11, 2008 #7
    y component of the velocity is zero and you still need to calculate the x-component, which you can use to calculate K.E. by using the formula, K.E. = 1/2 mvx2
  9. Oct 11, 2008 #8
    Vx always remains the same in a parabolic trajectory, so it will be same as initial velocity along x direction which will be 118 cos(41.2)
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