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I don't have a clue how to find the coordinates

  1. Jan 28, 2013 #1
    1. The problem statement, all variables and given/known data

    http://i277.photobucket.com/albums/kk63/lioricsilver/Untitled_zps7cf41f04.png


    2. Relevant equations

    y2=(16)x is the equation on the question


    3. The attempt at a solution
    I have got any clue. I do know how to solve motion in 2d. But since time is not given or the height is not given I don't know how to apply any of what i know to this question.

    But my hypothesis is that since the hyperbola has an equation given i should use it to find minnimum of the curve. But thats all i got. Sorry but i found all attempts a dead end befor i begin cuz there is so less values to work with
     

    Attached Files:

  2. jcsd
  3. Jan 28, 2013 #2

    SteamKing

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    What condition must hold when the melon hits the bank?

    If you were to find the path of the melon when it is launched horizontally with the given initial speed, could you calculate the x and y values of the trajectory, ignoring the shape of the bank?
     
  4. Jan 28, 2013 #3

    tms

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    The problem is not particularly well-worded. The vertex of the parabola must be the point where the melon takes off, as shown in the picture, not the edge of the road, as stated in the problem description. So the origin of the system is also as shown in the picture, and therefore the initial height is zero. (Of course, in general the actual location of the origin is arbitrary, but here it makes sense to locate it at the vertex of the parabola.)

    You have, or can get, an equation for the trajectory of the melon, and an equation for the stream bed. What must be true, as SteamKing asked, when the melon meets the ground?
     
  5. Jan 29, 2013 #4
    Ok after analyzing the data i got here is what i know and don't know


    Object horizontal stuff

    constant velocity V=10m/s
    Acceleration a=0
    Time t=?
    Distance s=?

    Vertical Stuff

    Initial Velocity U=0
    Final velocity V=?
    Acceleration = 9.81m/s2(10m/s2)
    Time t=?
    Distance s=?

    But i do know that the time taken for both is the same

    Know where do i go from here?
     
  6. Jan 29, 2013 #5

    tms

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    Start by writing down the relevant equations. Those would be the kinematics equations that relate position, velocity, acceleration, and time.
     
  7. Jan 29, 2013 #6
    I did try v=u+at for the vertical velocity

    so v=0+10 x t
    v=10t

    also for vertical i tried s=ut+1/2at2
    so s=0t+1/210t2
    s=0+5t2
    s=5t2

    also i tried v2=u2+2as
    so v2=0+2 x 10 x s
    v2=20s

    what do i do now????
     
  8. Jan 29, 2013 #7
    I also mixed up the first 2 equations

    v=10t and s=5t2 by substituting in the place of t and got
    v=10 x √ s / √ 5

    square all to get rid of root and got

    v2=100 x s/5

    substitute this with 3rd equation v2=20s

    and got 400s2= 100s/5

    so make subject s

    and by now i feel faint
     
  9. Jan 29, 2013 #8
    some one plz give me hand plz. I need some sort of guidance for this
     
  10. Jan 29, 2013 #9

    SteamKing

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    Reread post #2
     
  11. Jan 30, 2013 #10
    Ok here is some more info.

    that equation that is given holds the final key

    y2=16x

    so i guess i need to put distance in x and y even in terms of s and solve

    Plz tell me if this could work

    And i couldn't understand what steamking asked in post 2

    Well the question says the melon splatters on the bank
     
  12. Jan 30, 2013 #11

    SteamKing

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    What can we say about the x and y coordinates of the melon and the x and y coordinates of the bank when the melon hits the bank?
     
  13. Jan 30, 2013 #12

    SammyS

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    It looks to me as if the problem is worded just fine. It's the art work that doesn't fit the problem too well.
    No. The horizontal component of v is a constant: vx = 10 m/s .

    The x coordinate of the melon, as a function of time is
    x = vx∙t

    x = 10∙t .​

    Note: use the X2 icon for super scripts.

    You are almost correct regarding the vertical position of the melon. (Of course the melon's vertical coordinate goes down from zero, not up.)

    So the y component of the melon is
    y = -5∙t2
    .

    Solve x = 10∙t for t and plug that into the equation for y, to find the path taken by the melon.

    You now have two simultaneous equations in x and y to solve .
     
  14. Feb 2, 2013 #13
    Thank you very.
    I have got it by now the only thing that was bothering me was the equation of the parabola.
    But i got it now
     
  15. Feb 5, 2013 #14
    Could i post all my working and final answer so that you could tell me if it is right or wrong???
     
  16. Feb 5, 2013 #15

    SammyS

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    Yes, that is perfectly fine.
     
  17. Feb 6, 2013 #16
    ok here is the answer. plz check it out for me
     
  18. Feb 6, 2013 #17
    Distance in x axis
    s=ut+1/2at2

    s=? U=10 t=? a=0

    ∴ s=10t

    Distance in y axis
    s=ut+1/2at2

    s=? U=0 t=? a=9.8

    s=9.8/2 t2

    t in x axis and t in y axis is same

    substitute distance in x and y of y2= 16x

    y2= 16x

    (9.8/2 t2)2 = 16 x 10t

    24.01t4=160t

    make subject t

    24.01t4/t= 160

    24.01t3=160

    t3=160/24.01

    t3= 6.664

    t=3√6.664

    t=1.88s

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

    substitute t= 1.88s in equations in x and y axis

    s=10t

    10x 1.88

    =18.81

    s=9.8/2 t2

    9.8/2 (1.88)2


    x= 18.81
    y=17.35

    is this correct????

    I gave everything in very detail so you can tell me where i went wrong.
     
    Last edited: Feb 6, 2013
  19. Feb 6, 2013 #18

    SammyS

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    That's for the x coordinate of the melon.

    Therefore, x = 10t for the melon.

    That's for the y coordinate of the melon, except that it should be -9.8/2 t2.

    Therefore, y = -9.8/2 t2 for the melon.

    Solve x = 10t for t.

    Substitute that into y = -9.8/2 t2.

    That will give y in terms of t, for the melon's trajectory.

    Take this result together with the equation for the river bank.

    This gives two equations in two unknowns. Solve for x any y. (There are two solutions.)
     
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