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Find initial velocity given angle and distance to go over an object

  1. Jun 19, 2009 #1
    1. The problem statement, all variables and given/known data

    A baseball leaving the bat at 46 degrees at a height of 1.2m from the ground clears a 3m high wall 125 meters from home plate. what is the initial velocity of the ball?

    that's my problems i am have a reasonable amount of trouble with it. any help would be much appreciated.
     
  2. jcsd
  3. Jun 19, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi jarn6700! Welcome to PF! :smile:
    ok … use the usual constant acceleration equations, for the x and y directions separately, and remember that there is no horizontal force, so ahorizontal = 0. :wink:

    Show us how far you get, and where you're stuck, and then we'll know how to help! :smile:
     
  4. Jun 19, 2009 #3
    yeah i gather there is no acceleration in the x-axis and that there is the
    -9.81 acceleration in the y-axis. but i can't seem to figure out firstly the velocity to reach the 125m and then secondly how to account for the 3m high wall.

    i've had a read of some general initial velocity thread which were a bit confusing (i.e. https://www.physicsforums.com/showthread.php?t=314614&highlight=initial+velocity ) so anything to start us off is great.

    cheers
     
  5. Jun 19, 2009 #4
  6. Jun 19, 2009 #5

    tiny-tim

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    x goes from 0 to 125, and y goes from 0 to 3 …

    call the total time "t" …

    that should give you two equations, and then you can eliminate t.

    Show us how far you get, and where you're stuck, and then we'll know how to help! :smile:
     
  7. Jun 19, 2009 #6
    x-axis 125 = Vi x cos(46)t
    t = 125/Vi x cos(46)

    then do i sub that into
    [tex]
    y=-\frac{1}{2}g t^2+v_{y,0}t[/tex] ?

    0 = -4.9t^2 + vsin(46)t

    divide both sides by t getting

    0 = -4.9t + vsin(46)

    4.9t = vsin(46)

    4.9 = vsin(46)/t

    4.9 = 0.72v/125/0.7v
     
    Last edited: Jun 19, 2009
  8. Jun 19, 2009 #7

    tiny-tim

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    (try using the X2 tag just above the Reply box :wink:)

    Yup! :biggrin:
    Nooo :cry:, it's not 0, it's 3 - 1.2 …

    Try again! :smile:
     
  9. Jun 19, 2009 #8
    not sure what you are referring to with the x^2 tag (please explain :) )


    3 - 1.2 = -4.9t^2 + vsin(46)t

    divide both sides by t getting

    1.8/t = -4.9t + vsin(46)

    1.8/t + 4.9t = vsin(46)

    and now i am stuck. yup, very stuck.
     
  10. Jun 19, 2009 #9

    tiny-tim

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    (if you click the QUOTE button, you get the Reply page, and just above the text field there's a line that begins B I U … there's an X2 and X2 tag near the end, and if you click them it gives you supersripts and subscripts :wink:)

    why?? … this is a straightforward quadratic equation as it is! :smile:

    (but also why haven't you substituted the value for t from the horizontal equation, so as to give you a quadratic in vi?)
     
  11. Jun 19, 2009 #10
    1.8 = -4.9(125/0.7v)^2 + 0.72v(125/0.7v)

    0 = -875v^2 + 128.57v -1.8


    is that correctly put into the equation? sorry i battle with quadratics...
     
  12. Jun 19, 2009 #11

    tiny-tim

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    Yes (except you didn't square the 125 :rolleyes:)

    Now solve by using the (-b ±√(b2 - 4ac))/2a formula. :smile:
     
  13. Jun 19, 2009 #12
    whe you say i didnt square the 125: i thought you were meant to leave the (125/0.7v)^2 as a squared number to be able to put into the formula?

    -76564.5v + 128.7v - 1.8

    (-128.7 ±√(128.7)^2 - 4 (-76564.5)(-1.8))/2(-76564.5)
    (-128.7 ±√534700.71)/-153129
    (-128.7 ±731.23)/-153129
    that right so far?

    (did -875v^2 + 128.57v -1.8 just incase, wasnt sure.)
    (-128.7 ±√10263.69)/-1750
    (-128.7 ± 101.31)/-1750

    thanks for your help so far, been very useful
     
    Last edited: Jun 20, 2009
  14. Apr 12, 2011 #13
    can someone tell me where 0.7v in the last line came from? I get that .72v=vsin(46) and t=125 but i am lost as to where that 0.7v snuck in.....it also took me a while to figure out that the x value is time. I thought of it as distance not time. I was lost, but i am now found(sort of)
     
  15. Apr 13, 2011 #14

    tiny-tim

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    hi mlbuxbaum! :smile:

    from …
    (cos46° = 0.7) :wink:
     
  16. Apr 13, 2011 #15
    I figured that .72 was sin46°(actually .7193), but in the eq he shows
    4.9=vsin(46)/t
    4.9=.72v/125/.7v

    not sure why the equation is divided by cos46°.

    my problem is alittle more meaty than this original.(to me)
    This is for my Intro to Engineering class, for the Engineering with Excel portion. The instructor wants us to find initial velocity at 15,30,45 & 60°, create a chart(graph) for each angle by creating a chart of 11 x, y coordinates for each angle. He gave us a formula to use,
    y=y0+xtan[tex]\Theta[/tex]0-.5g(x2/v2cos2[tex]\Theta[/tex]0)
    Then he told us to set y=0 to get the range, and then use the quadratic equation to get the x.

    I really like how you help walk us through these problems. Just the answers dont help , i want to learn how to do it. I really hope you can help me with this. I am most likely over-thinking this and losing myself in the process.
     
  17. Apr 13, 2011 #16

    tiny-tim

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    hi mlbuxbaum! :wink:

    jarn6700 :smile: was solving the two simultaneous equations …


    "simultaneous" means that they have the same value of t,

    so we can eliminate t by solving for t in one equation, then substituting that value of t in the other equation

    the cos46° comes from the first equation :wink:
    i don't see the difficulty …

    you have 0 = y0+xtan[tex]\Theta[/tex]0-.5g(x2/v2cos2[tex]\Theta[/tex]0),

    you put in one given angle at a time, and use the standard formula for solving a quadratic equation (-b ±√(b2 - 4ac))
     
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