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Kinematics projectile motion with constant acceleration.

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

    According to the Guinness Book of World Records, the longest home run ever measured was hit by Roy “Dizzy” Carlyle in a minor league game. The ball traveled 188 m before landing on the ground outside the ballpark.

    Assuming the ball's initial velocity was 56 degrees above the horizontal and ignoring air resistance, what did the initial speed of the ball need to be to produce such a home run if the ball was hit at a point 0.9 m above ground level? Assume that the ground was perfectly flat.

    2. Relevant equations

    Any constant acceleration equations.

    3. The attempt at a solution

    Ok, so this is my problem. I have all the meters, but can't find any velocities.

    x = 188
    y= 0


    ay = -9.8
    ax=0

    Xo = 0
    Yo = 0.9

    ? sin 56 = Voy
    ? cos 56 = Vox

    I don't really know what to do from here. ='(

    I did write 0= Voy^2+2(-9.8)(-0.9) = 4.2 m/s = Voy, and I have a lot of values after that, but I don't think I'm doing it right.
     
    Last edited: Sep 24, 2011
  2. jcsd
  3. Sep 24, 2011 #2
    Write down the constant acceleration equations that you know of, then plug in the values you have for the terms. You need only one constant acceleration equation. The one for distance.
     
  4. Sep 24, 2011 #3
    0-0.9=Voy(t)+.5(-9.8)(t^2)

    Vy-Voy =(-9.8)(t)

    V^2y = V^2oy + 2(-9.8)(0-0.9)

    Is Voy 0?
     
  5. Sep 24, 2011 #4
    Voy is the component of the initial velocity in the y direction. That is, if the magnitude of the initial velocity is Vo, then it's Vo sin 56.

    Also, Vox = Vo cos 56.

    The three equations you have written down are correct. But you've considered only the y direction. Write down the same equations for the x direction.
     
  6. Sep 24, 2011 #5
    188-0=Vox(t)

    Vx = Vox

    V^2x = V^ox
     
    Last edited: Sep 24, 2011
  7. Sep 24, 2011 #6
    The 3 equations you've written for the x direction are also correct.

    But wait. You want to add something to your equations for the y direction? Why do you think 0= V^2oy- V^2y ?
     
  8. Sep 24, 2011 #7
    I think they equal zero because the velocity at he beginning will equal the velocity at the end, except it will be negative, thus giving me Voy=4.2 m/s.
     
  9. Sep 24, 2011 #8
    The equation of motion is,

    V^2y = V^2oy + 2*ay*(y-Yo)

    So,

    V^2y - V^2oy = 2*ay*(y-Yo)

    Which is not equal to 0. The velocity at a height of 0.9 m will be the negative of velocity at the beginning. Not the velocity at 0 height.

    OK, proceeding with the solution to the problem,

    You have for the y direction,

    0-0.9=Voy(t)+.5(-9.8)(t^2)
    Vy-Voy =(-9.8)(t)
    V^2y = V^2oy + 2(-9.8)(0-0.9)

    You have for the x direction,

    188-0=Vox(t)
    Vx = Vox
    V^2x = V^ox

    Also you have,

    Vox = Vo cos 56
    Voy = Vo sin 56

    Which ones of these 8 equations can you use to find the answer. Look for common variables between equations.
     
  10. Sep 24, 2011 #9
    Ok, thanks for the explanation.

    188= Vo(cos 56)t

    -0.9=Vo(sin 56)t
     
  11. Sep 24, 2011 #10
    Where did you get this one from?
     
  12. Sep 25, 2011 #11
    x-xo = Vo(cos 56)t
    188-0= Vo(cos 56)t

    y-yo = Vo(sin 56)t
    0-0.9= Vo(sin 56)t
     
  13. Sep 25, 2011 #12
    You had this equation before,

    0-0.9=Voy(t)+.5(-9.8)(t^2)

    How did the ".5(-9.8)(t^2)" term disappear?
     
  14. Sep 25, 2011 #13
    ok, it doesn't.

    Thanks.
     
    Last edited: Sep 25, 2011
  15. Sep 25, 2011 #14
    part b) How far would the ball be above a fence 3.0m high if the fence was 116 m from home plate?

    I received 45 m/s for initial velocity finally! thanks!

    I don't know where to start part b.

    nevermind, i did it.
     
    Last edited: Sep 25, 2011
  16. Sep 25, 2011 #15
    Do you mean you did part b) ?
     
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