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Motion in Two Dimensions (Vectors)

  1. Sep 3, 2007 #1
    1. The problem statement, all variables and given/known data

    One of the fastest recorded pitches in major league baseball, thrown by Billy Wagner in 2003, was clocked at 101.0 mi/hr. If a pitch were thrown horizontally with this velocity, how far would the ball fall vertically ( in feet ) by the time it reaches home plate, 60.5 feet away?

    2. Relevant equations

    T1 = Delta(X)/ V1
    Delta (Y) = Vo T + (1/2) A T^2

    3. The attempt at a solution

    Delta (Y) = (0)t + (1/2) (-21.9 mi/hr) (.000099 hr)
    Delta (Y) = -1.07 * 10^-7 mi OR 5.66 * 10 ^-4 FT

    This answer doesn't seem correct at all, any help?
  2. jcsd
  3. Sep 3, 2007 #2


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    Well, without looking too hard at this:

    Are you sure your units are correct? Your first calculation has miles and hours, but then you state your answer in feet.
  4. Sep 3, 2007 #3
    Thats what I found so difficult about the problem, was all of the conversions between units... Would it be easier to solve for everything in miles and then muliply times 5290 feet after the solution? Im really confused with this one.
  5. Sep 3, 2007 #4


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    I personally would convert everything into feet and seconds, since g = 32.2 ft/s^2 (I think, I'm used to metric). But it doesn't really matter, as long as everything is consistent it should work either way.

    So try to convert 101.0 mi/hr into ft/s. That should make everything easier.
  6. Sep 3, 2007 #5
    How can I go about finding time, because V=Vi + at doesn't work with what Im given???
  7. Sep 3, 2007 #6


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    The time for the ball to reach the plate? You use the first equation you listed up under Relevant Equations, it is correct since the velocity of the ball in the horizontal direction remains constant. I think your method is OK, you just need to get the units right. Are you stuck on the conversion?
  8. Sep 3, 2007 #7
    The Vinitial(X) and Vfinal(X) are 148.4 ft/s, correct? If so, I think I have it...

    .41 Sec = t

    How would I go about solving for the velocity?
  9. Sep 3, 2007 #8


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    I don't understand, you already have the velocity. :confused:

    All there is left to do is find the distance the ball drops in the vertical direction as it travels the 60.5 ft.

    I've noticed there is an error in your initial calculation for delta(y), you forgot to square the time term.
  10. Sep 3, 2007 #9
    Don't I need a separate velocity for the Y component of the problem?
  11. Sep 3, 2007 #10


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    No, because you've been given enough information that you don't need it. You know the initial vertical velocity is zero, since the ball was released horizontally. You have found the time of travel, so you know how long the ball is falling for as it travels. You also know the acceleration it is experiencing due to gravity. Looking at your second equation, we see that you have the numbers for all of the variables present. So you can solve it to find the distance the ball dropped.
  12. Sep 3, 2007 #11
    okay, thanks! So here is what I have:

    Delta (Y) = Vinitial(Y) (t) + (1/2) g t^2

    = 0(t) + (1/2) (-9.8) (.41^2)

    Delta (Y) = 2.71 ft

    Is this correct?

    Thanks for the great help!
  13. Sep 4, 2007 #12


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    That looks reasonable to me, but be careful of your signs. You could have use g = 32.2 ft/s^2 to get the answer directly in feet, and saved yourself a step. Eh, works either way.
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