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A ball is hit with initial velocity -- Find the Horiz/Vert displacement

  1. Mar 16, 2017 #1
    1. The problem statement, all variables and given/known data
    A softball is hit with an initial velocity of 29 m/s. at an angle of 60 degrees above the horizontal and impacts the top of the outfield fence 5s later. Assuming the initial height of the softball was .5 m above (level) ground, what are the balls horizontal and vertical displacements?

    2. Relevant equations
    [itex] a_xt + V_i = V_f [/itex]
    [itex] \Delta x = (1/2) (V_f + V_i ) t [/itex]
    [itex] \Delta x = V_it + (1/2)a_xt^2[/itex]
    [itex] \Delta x = V_f - (1/2) a_xt^2 [/itex]

    3. The attempt at a solution
    [itex] V_y = 29sin(60) = 25.11473671 m/s[itex]
    [itex] V_x = 29cos(60) = 14. 5 m/s [/itex]
    Horizontal displacement is easy. Using formula [itex] \Delta x = (1/2) (V_f + V_i ) t [/itex] with [itex] V_f = V_i [/itex] you get [itex] \Delta x = V_ft[/itex], so horizontal displacement = (14.5 m/s)(5s) = 72.5 m

    Now to find vertical displacement:

    I use formula [itex] a_xt + V_i = V_f [/itex] to find final velocity, plugging in I get:
    [itex] (-9.8 m/s^2) (5s) + 25.11473671 m/s = V_f [/itex]
    = [itex] -23.88526329 m/s = V_f [/itex]

    Now using [itex] \Delta y = (1/2)(V_f + V_i)(t) [/itex] I get:

    [itex] \Delta y = (1/2)(-23.88526329 m/s + 25.11473671 m/s) (5s) [/itex]

    = 3.07368355 m = y displacement

    But the softball starts off with in initial height of .5 m, so I subtract y displacement - .5, which =

    2.57368355 m,

    but my book gives me answer: 2.95 m

    Can anyone tell me where I went wrong here?? I don't see my mistake..
  2. jcsd
  3. Mar 16, 2017 #2


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    Staff: Mentor

    You're taking several differences of numbers that are fairly close in magnitude. You are keeping lots of extra digits to maintain accuracy, which is good, however it doesn't make sense to use such accuracy if one of your starting values has only one decimal place of accuracy. I'm referring to the value of g that you're using.

    Try using a more accurate value for g, such as ##g = 9.80665~m/s^2##.

    Alternatively do all the work symbolically first, only plugging in numbers when you've reduced the expression and removed all redundant operations. It's a less error prone way to work, avoiding rounding and truncation issues as well as transcription errors from carrying along all those digits through the algebra.
  4. Mar 16, 2017 #3
    I have tried using that value of g and there was an even bigger difference from the books answer to the one I got...

    I'm not really sure where the error has come from after I've done it algebraically as well... Does my method look good to you?
  5. Mar 16, 2017 #4


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    The ##\Delta y## you calculated is the y-displacement. There is no need to subtract ##0.5m##.

    However, to echo what @gneill says, your use of numbers to umpteen decimal places is clumsy, makes your work difficult to read and even more difficult to spot an error, and is technically wrong (given you do not know the input data to this degree of accuracy).

    Finally, it would have been simpler to get both displacements from:

    ##\Delta x = v_x t## and ##\Delta y = v_yt - \frac12 gt^2##
  6. Mar 16, 2017 #5
    I think the book used g = 9.81 m/s^2, instead of 9.8 m/s^2.
  7. Mar 16, 2017 #6
    Why is there no need to subtract .5 m? Isn't the formula [itex] \Delta y = V_yt - \frac 12 gt^2 [/itex] assuming that it starts at level with the ground?
  8. Mar 16, 2017 #7
    Δy means the change in y, or the displacement in the y direction. The displacement will always be the same, regardless of the starting position. The problem statement asked for the horizontal and vertical displacements - not position.
  9. Mar 17, 2017 #8


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    No, that equation makes no assumption about a starting height, because it calculates ##\Delta y##.

    In fact, it would have made more sense to add the ##0.5m##, which would have given you the height of the ball.
  10. Mar 17, 2017 #9
    Ahh alright. I got you guys. I understand now, displacement formula = displacement, lol. Doesn't matter where it starts.

    I got pretty close to the answer, with the answer being 2.94868355

    Apparently I need to use the full value for initial y velocity (25.11473671) and g = 9.81
  11. Mar 17, 2017 #10


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    Do you think you can calculate the position of a softball to the nearest nanometer?
  12. Mar 17, 2017 #11
    No, so my answer would have to be rounded to 2.95 meters! Which is the same as my books answer.
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