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Height when a ball is thrown vertically at half its velocity

  1. Jan 31, 2014 #1
    [itex][/itex]1. The problem statement, all variables and given/known data
    A baseball is thrown vertically into the air with a velocity, v, and reaches a maximum height, h. At what height was the baseball with one-half its original velocity? Assume no air resistance.


    2. Relevant equations
    v22 = v12 + 2ad ?



    3. The attempt at a solution
    let 10m/s be the original velocity
    d = [itex]\frac{v^{2}_{2} - v^{1}_{2}}{2g}[/itex]
    d = [itex]\frac{0 m/s - (10 m/s)^{2}}{2(-9.8 m/s^{2}}[/itex]
    d = 5.1

    at 5 m/s:


    d = [itex]\frac{0 m/s - (5 m/s)^{2}}{2(-9.8 m/s^{2}}[/itex]
    d = 1.28

    [itex]\frac{1.28}{5.1}[/itex] = 0.25

    but the answer is 0.75 of the original height. How to solve this?
     
  2. jcsd
  3. Jan 31, 2014 #2

    Dick

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    The d you are calculating is the distance between the point where the velocity v=0 and the point where the velocity is v/2. The point where v=0 is at the TOP of your trajectory.
     
  4. Jan 31, 2014 #3
    I didn't fully understand what you mean
     
  5. Feb 1, 2014 #4

    NascentOxygen

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    Staff: Mentor

    Ace, in the second part are not using the correct reference level.

    The question you should be answering is: if a ball is thrown vertically upwards, at what height above the ground has its velocity dropped back to half of what it initially had?
     
    Last edited: Feb 1, 2014
  6. Feb 1, 2014 #5

    Dick

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    You are calculating distances from the point where v=0. So the second distance you calculated isn't the distance from the ground. It's distance from the point where v=0. v=0 is at the top of the trajectory.
     
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