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Somebody please help me

  1. Nov 29, 2004 #1
    Somebody please!!!! help me

    I have already posted this question but no one has given me any help. I need this answer, it is due soon.

    A daredevil is shot out of a cannon at 24.0° to the horizontal with an initial speed of 26.0 m/s. A net is positioned a horizontal distance of 50.0 m from the cannon. At what height above the cannon should the net be placed in order to catch the daredevil?

    I know how to set it up... I think

    Vx= Vcos(theta)
    Vy=Vsin(theta)

    x = Vxt
    t = x/Vx

    y = Vyt + .5gt^2

    HELP ME
     
  2. jcsd
  3. Nov 29, 2004 #2
    Nope -

    y = Vyt - .5gt^2

    Vy and g are in opposite directions

    So you know t and you know how high the daredevil is at t
     
  4. Nov 29, 2004 #3
    When I do that I get 43.97... which is wrong. Can you tell me what I've done wrong?
     
  5. Nov 29, 2004 #4
    If you are using y = Vyt - .5gt^2 make sure you are using 9.80 m/s^2 for g. The minus sign accounts for what some textbooks give for gravity, 9.80 m/s^2. Only use y = Vyt + .5gt^2 when you are using -9.80 m/s^2 for gravity. Be sure not to mix them up.

    z-component
     
  6. Nov 29, 2004 #5
    I was taught Vertical components have nothing to do with horizontal components, so here is how i would solve it.

    A daredevil is shot out of a cannon at 24.0° to the horizontal with an initial speed of 26.0 m/s. A net is positioned a horizontal distance of 50.0 m from the cannon. At what height above the cannon should the net be placed in order to catch the daredevil?

    Start with a freebody diagram and then solve for time in the X-Direction

    Xf = Xi +ViT + .5gT^2
    T^2 = Xf - ViT / .5g
    T^3 = Xf - Vi / .5g
    T^3 = 50 - 26 / .5(9.8)
    T^3 = 24 / 4.9
    T^3 = 4.9
    T = 1.7 seconds <--- Remember this

    Yf = Yi +ViT +.5gT^2 G is going to equal sin(24)9.8 due to the angle launched

    Yf = ViT + .5(cos(24)(9.8))T^2
    Yf = (26)(1.7) + .5(cos(24)(9.8))(1.7^2)
    Yf = 44.2 + 12.9
    Yf = 57.1 m above the end of the cannon barrel

    I'm not positive if that is correct or not, but it sounds reasonable to me

    I hope that helped you understand the 2D motion
     
  7. Nov 29, 2004 #6
    Well you added instead of subtracting. I get the surprising answer of 0.547 meter from the floor -- I would have thought it was going to be higher.

    {Vyt = 10.58*2.11 = 22.26 ; .5*g*t^2= 4.9*2.11*2.11 = 21.71. If you add instead of subtract you get your answer.}
     
    Last edited: Nov 29, 2004
  8. Nov 29, 2004 #7
    wow, those are 3 very different answers...can you explain how you got .57 meters? Because i may have showed him wrong :confused:
     
  9. Nov 29, 2004 #8
    I don't know how you guys got either of those answers

    I got 43.8
     
  10. Nov 29, 2004 #9
    Uhh!

    Xf = Xi +ViT + .5aT^2

    Assume no AR, a = 0, not 9.8
     
  11. Nov 29, 2004 #10
    I've gotten the 43.97 answer and the really small one, depending on whether you subtract or add the second part. Someone please help. It's due tomorrow!
     
  12. Nov 29, 2004 #11
    oh..no wonder i messed up..im not sure ill have to re-work this one
     
  13. Nov 29, 2004 #12
    Alright, I will explain how I did it.

    HORIZONTAL COMPONENT

    [tex]dD = 50 m[/tex]
    [tex]v_o = 24cos(30) m/s[/tex]
    [tex]v_f = ?[/tex]
    [tex]dT = ?[/tex]
    [tex]a = 0 m/s^2[/tex]

    We don't care about [tex]v_f[/tex].

    Solving for [tex]dT[/tex]:

    [tex]dD = v_i*dT+0.5*a*dT[/tex]
    [tex]50 = 26cos(24)*dT[/tex]
    [tex]dT = 2.1 s[/tex]


    VERTICAL COMPONENT

    [tex]dD = ?[/tex]
    [tex]v_i = 26sin(24) m/s[/tex]
    [tex]v_f = ?[/tex]
    [tex]dT = 2.1 s[/tex]
    [tex]a = -9.8 m/s^2[/tex]

    Solving for dD

    [tex]dD = v_i*dT + 0.5*a*dT^2[/tex]
    [tex]dD = 26sin24*2.1+0.5*-9.8*2.1^2[/tex]

    [tex]dD = 0.598[/tex]
     
    Last edited: Nov 29, 2004
  14. Nov 29, 2004 #13
    is there a website your doing all that on? or how do you get that font type/style
     
  15. Nov 29, 2004 #14
    is it me going crazy or is this a double post? :confused:
     
  16. Nov 29, 2004 #15
    Has anyone else used LON-CAPA for homework before? I'm getting the exact same answers you guys are, but it says I'm wrong.
     
  17. Nov 29, 2004 #16
    Im sorry but what is this LON-CAPA?
     
  18. Dec 1, 2004 #17
    How high iis the cannon above the ground? Is that stated in the problem? If the height is H then the answer that many are getting should be incremented by H. The basic Newtonian setup has to be right.
     
  19. Dec 1, 2004 #18
    it probably has something to do with rounding. Because if you round the time to 2.1 seconds you get .598 m, but if you dont round it at all you get 0.548. Either way, the process is right, so play around with rounding.
     
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