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Homework Help: X and y component problem?

  1. Feb 12, 2008 #1
    [SOLVED] x and y component problem?

    1. The problem statement, all variables and given/known data

    http://loncapa3.physics.sc.edu/cgi-bin/plot.png?file=dengjh_sc_1202847653_17793457_plot.data [Broken]

    u will need that graph to solve this problem

    ) A golf ball is struck at ground level. The speed of the golf ball as a function of the time is shown in the figure., where t = 0 at the instant the ball is struck. How far does the golf ball travel horizontally before returning to ground level? i found this to be 112m and it is right

    What is the maximum height above ground level attained by the ball?


    2. Relevant equations
    S = (1/2)*a*t^2


    3. The attempt at a solution

    Looks like the ball travels for 4 sec. When the speed is minimum (that speed looks like 27.5 m/s to me) it has reached its peak so the vertical component of speed has gone to zero. All the remaining velocity is horizontal, 27.5 m/s. And that's the horizontal velocity,Vh, during the entire 4 seconds. So the horizontal distance h is found from
    Vh = h / total time

    Vertical distance: It took 2 seconds to return to the ground from that peak,
    S = (1/2)*a*t^2

    now i tried using the equation but it didnt work. can someone point me to the right directioin
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Feb 12, 2008 #2

    Hootenanny

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    Welcome to PF Lacy,

    Your URL is broken.
     
  4. Feb 12, 2008 #3
    thank you i fixed it
     
  5. Feb 12, 2008 #4

    Hootenanny

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    The reason it didn't work is your equation is incomplete. The full equation is,

    [tex]x = x_0 + v_0 t + (1/2) a t^2[/tex]

    Where x0 is the initial displacement (in this case zero) and v0 is the initial velocity.
     
  6. Feb 12, 2008 #5
    now im trying to find max height would i use y=vot-.5at^2

    would this be correct

    y=34(2)-(4.9)(4)

    48.4
     
    Last edited: Feb 12, 2008
  7. Feb 12, 2008 #6

    Hootenanny

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    Correct, where a = 9.81m.s-2 :approve:
     
  8. Feb 12, 2008 #7
    would this be correct

    y=34(2)-(4.9)(4)

    48.4
     
  9. Feb 12, 2008 #8

    Hootenanny

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    How did you determine the initial velocity?
     
  10. Feb 12, 2008 #9
    i looked on the graph and it started at 34m/s
     
  11. Feb 12, 2008 #10
    i also got this equation but its different from the first one

    you rearrange y-yo = Vosin(theta)t-
    1/2gt^2 so that sin(theta) = (1/2gt^2)/Vo. Once you have
    sin, plug in the variable in the y-yo = Vosin(theta)t-
    1/2gt^2 formula.

    which way do i do it im confused
     
  12. Feb 12, 2008 #11

    Hootenanny

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    That is the initial speed, not the initial [vertical] velocity. However, you know the horizontal velocity (112/4) and the speed, therefore you can easily calculate the initial vertical velocity.
     
  13. Feb 12, 2008 #12
    y=28(2)-(4.9)(4)

    alright now i got 36.4 is this right
    where vo=28 and t=2
     
  14. Feb 12, 2008 #13

    Hootenanny

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    [COLOR="#black"]White space[/COLOR]
     
  15. Feb 12, 2008 #14
    would this be it

    vo=.5at^2

    where a=9.8 t=2

    so vo=19.6m/s ???
     
  16. Feb 12, 2008 #15

    Hootenanny

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    You've use the wrong equation, but got the correct answer. The correct method would be as follows,

    [tex]v = v_0 + at[/tex]

    At maximum height [itex]v=0[/itex], hence,

    [tex]v_0 + at = 0 \Rightarrow v_0 = -2g = 19.6 m.s^{-2}[/tex]
     
  17. Feb 12, 2008 #16
    alright i got the right answer THANK YOU for all the help hootenanny :)
     
  18. Feb 12, 2008 #17

    Hootenanny

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    A pleasure Lacy :smile:

    If you'd be so kind to mark the thread as 'solved' (look for "Thread Tools" in the top right-hand corner of the thread) that would be great.
     
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