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Homework Help: Finding Initial velocity given certain variables

  1. Sep 19, 2010 #1
    1. The problem statement, all variables and given/known data

    To Find initial velocity, given final velocity, and displacement on Y axis.

    Hello. I've spent four hours trying different solutions to this following problem, none of which have worked. I've just started my undenominated science degree, and since I havn't started lectures yet I've worked ahead in my physics book, to avoid getting stuck later on in the year. Help with workings would be much appreciated.





    Given:

    "an baseball has velocity of 36 m/s, at an angle of 28 from the horizontal (x axis). The positive directions are to the right of the horizontal, and the top of the vertical axis. Ignoring air resistance, find the magnitude and direction of the initial velocity of the object."

    - The objects final position is displaced +7.5 metres on the Y axis from its original position. -- Final velocity is 36 m/s, 28 degrees down from the horizontal.
    - Acceleration on the X axis is obviously = 0m/s
    - Acceleration on the Y axis is = -9.80 m/s

    Unknown

    The Unknown variables are the:

    Magnitude of initial velocity.
    Direction of initial velocity.

    2. Relevant equations

    These are the equations I imagine are relevant.

    Initial Velocity

    Magnitude = Vo = √Vox2 + Voy2

    Direction = Theta = Tan inverse (Voy/Vox)

    3. The attempt at a solution

    Sorry, I'm not great with using the mathmatical symbols on this board. I'll do my best.

    V = 36m/s
    Vx and Vy form a right angle triangle to V.

    36/Sin 90 = Vy/ Sin 28

    36 m/s (Sin 28) = Vy

    Vy = 16.90

    362 = 16.92 + 19.12

    Vx = √19.1 = 4.37

    - - - - - - - - - - - -

    To find Voy, I tried oy = + √(-36 sin 28)2 - 2 (-9.80)(7.5) = 20.8

    Voy = 20.80 m/s
    Vox = 4.37 m/s

    - - - - -- - - - - - - - - - - -

    Problem:

    using pythagorases theorum for a right triangle, I found that the velocity V0 was = 21.25, but my book says that the magnitude for the velocity should be:

    38m/s

    Where am I going wrong?? I would really appreciate help, very sorry if I didn't provide sufficient workings but I think I did. And sorry if my inevitably idiotic error is frustrating for you to see, I'm not very mathematically inclined, which is the reason I'm practicing so much. Thanks

    Ryan
     
    Last edited: Sep 19, 2010
  2. jcsd
  3. Sep 19, 2010 #2

    Delphi51

    User Avatar
    Homework Helper

    I agree with your 16.9 for the Vfy, but I get
    Vfx = 36*cos(28) = 31.79 instead of your 4.37.
    You don't say how you got the initial speed. I used
    d = Vit+½at² and Vf = Vi+at simultaneously to get the vertical Vi, which resulted in a combined initial speed of 38.
     
  4. Sep 20, 2010 #3
    Thanks, I've redone the calculation and managed to get the Vfx and Vfy values correct. I've tried the d = Vit+½at² equation before, and got:

    Code (Text):
    7.5m = (V[SUB]i[/SUB])(t) - 4.9 (t)[SUP]2[/SUP].
    to find time I rejigged

    Code (Text):
    a = 0.5 (V[SUB]ox[/SUB] + V[SUB]t[/SUB])t
    to give

    Code (Text):
    t = a/0.5(V[SUB]ox [/SUB]+ V[SUB]x[/SUB])

    2t = (2)(-9.80)/(X + 16.9m/s)
    But their is both an X and T variable in this equation, so I have no idea how to solve for Vox.

    - - - - - -

    I also tried:

    Code (Text):
    V[SUB]y[/SUB][SUP]2[/SUP] = V[SUB]oy[/SUB][SUP]2[/SUP] + 2 (a[SUB]y[/SUB])(y)

    31.8[SUP]2[/SUP] = V[SUB]oy[/SUB][SUP]2[/SUP] + 2(-9.80)(7.5)

    31.8 = √voy[SUP]2[/SUP] - √147

    31.8 + 12.1 = V[SUB]oy[/SUB]

    43.9 m/s = V[SUB]oy[/SUB]
    the initial velocity is only 38m/s, and pythagoras's theorem implies that Voy < Vo. So this solution is false. Where am I going wrong.

    I appreciate the help, thanks.

    - - - - - - - - - - - - - -
     
    Last edited: Sep 20, 2010
  5. Sep 20, 2010 #4

    Delphi51

    User Avatar
    Homework Helper

    This looks good. You also have
    Vf = Vi + at
    16.9 = Vi - 9.8t
    If you solve the system of these 2 equations, you'll get Vi (vertical).
     
  6. Sep 26, 2010 #5
    To solve

    ½(-9.80)(t) + (vi)(t) = 7.5m
    vi = (16.9-9.8t)

    -4.9t + (16.9-9.8t)(t) = 7.5m

    quadratic equation

    9.8t² + 12t – 7.5 = 0

    {-b ± (square root)b² - 4ac} / 2a

    - 12 ± (square root) 144 – 4 (9.8)(-7.5) / 2(9.8)

    {12 ± 20.9}/19.6

    1.6 or – 0.5

    Vi = (16.9 – 9.8t)

    Vi = (16.9 – 9.8 (1.6)) < = > 16.9 – 9.8 (1.6)) ≤ 32 m/s

    Vi = 1.2

    ------

    Looking at the quadratic equation, I reckon the solution for t lies somewhere around 0.15s, which is definitely too small a time period to be correct.

    Gravity acts at -9.8m/s² in this case, and if the equation only ran for 0.5 seconds the ball wouldn’t even begin to move towards the negative y, given the initial velocity is 42m/s. It’d probably take at least take 3 seconds for the Vy = 0.

    The equation 9.8t² + 12t – 7.5 = 0 is obviously, non – linear, since when this thing essentially describes a differential curve, but I don’t now how to solve non linear equations. Are they solved using matrixes or something??

    Sorry for wasting so much time with this stupid question, but it is pretty important that I learn how to derive Vo’s from V’s, and visa versa, it’s the fundamental element of kinematics really. I would trouble my lecturers, but we havn’t even started Vectors yet in Physics, and in a class of 400 they hardly want to be bothered by a student.

    -------
     
  7. Sep 26, 2010 #6

    Delphi51

    User Avatar
    Homework Helper

    You have 7.5m = (Vi)(t) - 4.9 (t)² [1]
    and Vf = Vi + at
    -16.9 = Vi - 9.8t [2]
    Solve [2] for Vi = 9.8t - 16.9 and sub in [1]:
    7.5 = 9.8t² - 16.9 t - 4.9t²
    0 = 4.9t² - 16.9t - 7.5
    Once you get t, you can use [2] again to find the vertical Vi.
     
  8. Oct 13, 2010 #7
    Solved it. Thanks a million for the help, I really appreciate it:D
     
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