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Physics Projectile Motion Question

  1. Aug 19, 2010 #1
    1. The problem statement, all variables and given/known data

    The question is: A fireman at point A wishes to put out a fire at B. Determine the two possible angles [tex]\theta[/tex]1 and [tex]\theta[/tex]2 at which this can be done. The water flows from the hose at vA=28m/s and the value for gravity is 9.81m/s2

    [PLAIN]http://users.adam.com.au/shortround/pm.JPG [Broken]

    2. Relevant equations

    [tex]
    v = v_0 + a t
    [/tex]

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

    [tex]
    v^2 = v_0^2 + 2 a \Delta x
    [/tex]

    3. The attempt at a solution

    We need to seperate vA into the horizontal and vertical components. But to do this we need to know either [tex]\theta[/tex] or one of the components, but we do not know either.

    I have spent a long time on this question and have not got very far. I have tried using trial and error, but I've had no luck. I know that there must be a better way to do it. The angle could be above or below the horizontal. From memory the two angles must be complementary, but I'm not entirely sure on that.

    If anyone could just give me a couple of hints to get me going, I would be very grateful.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 19, 2010 #2
    Ok, first thing you are trying to find the 2 angles. So you can't know them. But you can solve the question using algebraic equations. (Sorry I am new to this so I may not be really clear)

    But you do know that one component of the velocity will be (the horizontal one) will be V (velocity) times the cosine of the angle. The vertical component will be the the velocity times the sinus of the angle. There you go you now have divided the speed into its components...
     
  4. Aug 19, 2010 #3
    you know every other variable except for the sinus and the co-sinus, so you can solve for (by plugging the components into the relevant equations) them and hopefully get the angle by using arcsin(value of sinus).
     
  5. Aug 19, 2010 #4
    I understand that the initial horizontal velocity is 28cos[tex]\theta[/tex] and the vertical component is 28sin[tex]\theta[/tex].

    As iraten said I could then substitute it into [tex]

    v^2 = v_0^2 + 2 a \Delta x

    [/tex], but we do not know the final velocity when the water hits the wall. In the horizontal direction a=0 and in the vertical direction a=9.81m/s2. But I am still unsure on how all of this comes together. Any more help would be great!
     
  6. Aug 19, 2010 #5
    you can use the energy equation to find the final velocity: mgh = 0.5mv^2, write down all the equations you can, then try to find a way to isolate sin (remember sin^2+cos^2 = 1)
     
  7. Aug 19, 2010 #6
    In my text book they have an example where the initial velocity is 24m/s and they give the answer as [tex]\theta[/tex]=24.8[tex]o[/tex] below the horizontal and [tex]\theta[/tex]=85.1[tex]o[/tex] above the horizontal.

    Okay using [tex]v^2 = v_0^2 + 2 a \Delta x[/tex] I found the final velocity in the y direction to be: ((24sin[tex]\theta[/tex])2+(2*(-9.81)*6))1/2 = ((24sin[tex]\theta[/tex])2-117.72)1/2

    By putting this back into [tex]v = v_0 + a t[/tex], with v0=24sin[tex]\theta[/tex], I found the time was t=1.106 seconds. Finally I put this into [tex]x = v_0 t + (1/2) a t^2[/tex] and got a value of [tex]\theta[/tex]=51.57[tex]o[/tex]. But this was not the value given by the text book.

    What have I done wrong?

    Also iratern, thanks for your help. I am slowly getting there. I have been doing much harder problems than this one fine. I don't know why I just can't get this one :S
     
    Last edited: Aug 19, 2010
  8. Aug 19, 2010 #7
    sorry I haven't worked out the problem, I would but it's late here so I am on the sleepy side... But it's probably in the algebra or in the simple math keep on it...
     
  9. Aug 19, 2010 #8

    rl.bhat

    User Avatar
    Homework Helper

    In the problem x, y, vo and g is given. You have to find θ.
    You know that
    x = vo*cosθ*t ..(1) and
    y = vosinθ*t - 1/2(g)t^2....(2)
    So t = x/(vo*cosθ) and
    y = vosinθ*x/vocosθ - 1/2(g)(x/vocosθ)^2
    y = x*tanθ -1/2g*x^2*sec^2(θ)/vo^2
    y = x*tanθ -1/2g*x^2*[1 + tan^2(θ)]/vo^2

    Substitute the values and solve for tanθ.
     
  10. Aug 20, 2010 #9
    Thanks rl.baht! I had just managed to work it out using the same method :)
     
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