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Projectile Motion homework problem

  1. Sep 25, 2007 #1
    A projectile is fired from the surface of level ground at an angle Ao above the horizontal.
    A) Show that the elevation angle B of the highest point as seem from the launch point is related to Ao by tanB = (1/2)tanAo
    B) Calculate B for Ao = 45 deg

    I figure at the highest point, Vy must equal 0.
    let H be height of launch
    let R by range of launch

    so tanB = H/(1/2)R
    and tanB = (1/2)tanAo

    H/(1/2)R = (1/2)tanAo
    tanAo = H/(1/4)R

    I'm not sure how to proceed, do I want to derive H or R in terms of velocity/acceleration?

  2. jcsd
  3. Sep 25, 2007 #2


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    I'm confused... how did you do this last step? This is what you're supposed to prove, but you haven't proven it...

    Find H and R in terms of Ao... then you can substitute them into tanB = H/(R/2)
  4. Sep 25, 2007 #3
    I was just stating what it has to equal, I guess I can't do that since that is what I'm trying to prove

    Can I use the statement: tanB = (1/2)tanAo to prove it?
    for instance
    H/(1/2)R = (1/2)tanAo
    tanAo = H/(1/4)R

    H = (1/4)R*tanAo
    R = H/(1/4)tanAo
  5. Sep 25, 2007 #4


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    no. you can't use the statement that you're trying to prove...

    forget about B for now... try to get H and R in terms of Ao.... suppose you have an object with initial velocity v at an angle Ao... now what is the height in terms of v and Ao? What is the range in terms of v and Ao?
  6. Sep 25, 2007 #5

    When Vy = 0
    the height will = H
    and the range will = (1/2)R

    since the opposite and adjacent side of tanAo wont be in terms of H yet
    let J = tanAo(1/2)R
    since the adjacent side of both angles will be equal

    for initial velocity

    Vox = VocosAo
    Voy = VosinAo

    velocity components:
    Vx = Vox
    Vy = Voy - gt

    for displacement
    x = Voxt
    y = Voyt - (1/2)gt^2

    y = (tanAo)x -[g/(2VocosAo)]x^2

    R = (2Vo^2)sinAocosAo / g
    R = (Vo^2)(sin2Ao) / g
    R = (2Vo^2)(cosAo^2)(tanAo) / g

    tanAo = gR / (2Vo^2)(cosAo^2)

    I'm getting lost here
  7. Sep 25, 2007 #6


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    I think you're making it more complicated than it needs to be...

    You know the time taken to get to maximum height is Vosin(theta)/g. So the time to hit the range is 2Vosin(theta)/g

    x = Vocos(theta)t

    R = Vocos(theta)(2Vosin(theta)/g = 2Vo^2sin(theta)cos(theta)/g (this is what you got, but I'm not sure how you got it... I wasn't able to follow).

    What is H?
  8. Sep 26, 2007 #7
    is there a simpler way that you would approach R?

    wont H just be tanAo x (1/2) R
    except that it is on a curve

    or H = Voyt - (1/2)gt^2

    but at H Vy = 0
    im sortof lost here
  9. Sep 26, 2007 #8
    Try this:

    This is somewhat a special case
    1 Vyf=Vyi+Ayt
    2 0=Visin(yourangle)-gt1
    3 t1=((Vi)/g)sin(yourangle)
    4 h=[(Visin(yourangle)]∗((Visin(yourangle))/g)-(1/2)g(((Visin(yourangle))/g))²
    5 h=((Vi²sin²(yourangle))/(2g))
    6 R=Vxit2=(Vicos(yourangle)2t1
    7 =(vicos(yourangle))((2Visin(yourangle))/g)=((2Vi²sin(yourangle)cos(yourangle))/g)
    8 From trigonometry sin2(yourangle)=2sin(yourangle)cos(yourangle)
    9 so R=((Vi²sin2(yourangle))/g)
    10 range will max out at 45 degrees
  10. Sep 26, 2007 #9
    "your angle" referring to the angle Ao (angle of the launch) as opposed to B (angle who's hypotenuse will touch the highest point)
  11. Sep 27, 2007 #10


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    no, H = tanB*(1/2)R

    plug in t = Vosin(theta)/g into

    H = Vosin(theta)t - (1/2)gt^2

    to get H...

    or better yet.. using the kinematics equation vf^2 = vi^2 + 2ad. you know a = -g, vf = 0, vi = Vosin(theta)

    what is d?
  12. Sep 27, 2007 #11
    H = Vosin(theta)Vosin(theta)/g - (1/2)g(Vosin(theta)/g)^2
    H = Vosin(theta)^2 / g - (1/2)Vosin(theta)^2 / g
    H = Vosin(theta)^2 - (1/2)Vosin(theta)^2 /g
    H = (1/2)Vosin(theta)^2 / g


    vf^2 = vi^2 + 2ad
    0 = Vosin(theta)^2 +2gd
    d = -(Vosin(theta)^2) / 2(-g)
    d = (1/2)Vosin(theta)^2 / g

    can I use these to compare the 2 initial equations?
  13. Sep 27, 2007 #12


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    Now you know H and R...

    R = 2Vo^2sin(theta)cos(theta)/g
    H = (1/2)Vosin(theta)^2 / g

    substitute them into tanB = H/(R/2)

  14. Sep 27, 2007 #13
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