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Basic projectile question

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data

    A major leaguer hits a baseball so that it leaves the bat at a speed of 30.1 m/s and at an angle of 37.2 degrees above the horizontal. You can ignore air resistance.

    a) At what two times is the baseball at a height of 10.7 m above the point at which it left the bat?

    Voy = 18.2 m/s (30.1 sin37.2)
    Vox= 24.0 m/s (30.1 cos37.2)

    ax=0
    ay= -9.8 m/s^2

    yo = 0
    xo = 0

    vy= -18.2 m/s
    xy= 24.0 m/s


    2. Relevant equations

    any constant acceleration equations.


    3. The attempt at a solution

    I received Ttotal = 3.71 seconds which I think is right.

    I set up the equation to find time by doing 0-10.7 = 18.2t + .5(-9.8)t^2

    which is -4.9t^2 + 18.2t + 10.7

    I did the quadratic equation but received the times -0.516 and 4.23, and added together equal my total time of 3.71, however, I don't understand how there is a negative number.

    Are these the right times?

    P.S (this is a 7 part question, so I might need help after this question).

    Calculate the vertical component of the baseball's velocity at an earlier time calculated in part (a).

    I'm guessing this means use .732 seconds for time.
    I used the equation: vy=voy +ayt

    Vy= 18.2+(-9.8)(.732)

    =11.03 m/s.

    Is this right?
     
    Last edited: Sep 24, 2011
  2. jcsd
  3. Sep 24, 2011 #2
    Hey!
    Can you please explain how you arrived at substracting your desired height from zero?
    It should be, just 10.7, nothing else.
    Daniel
     
  4. Sep 24, 2011 #3
    Remember: Vy = Voy - gt

    I think you should revise your Vy => Although for this problem it is not necessary

    use this
    Yf = Yo + Voyt - .5gt^2

    just think where you want your Yf to be and solve for t
     
  5. Sep 24, 2011 #4
    I received the new times of 0.732 and 2.98 which sounds much better!

    I should have had 10.7-0, not 0-10.7.
     
  6. Sep 24, 2011 #5
    sounds much easier than my method! thanks.
     
  7. Sep 24, 2011 #6
    Always look for that blasted minus sign!
    Well done!
    Daniel
     
  8. Sep 24, 2011 #7
    thanks!
     
  9. Sep 24, 2011 #8
    Calculate the vertical component of the baseball's velocity at an earlier time calculated in part (a).

    I'm guessing this means use .732 seconds for time.
    I used the equation: vy=voy +ayt

    Vy= 18.2+(-9.8)(.732)

    =11.03 m/s.

    Nevermind, I know this is right.
     
    Last edited: Sep 24, 2011
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