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Homework Help: Projectile Motion Angles Question

  1. Jun 11, 2010 #1
    1. The problem statement, all variables and given/known data

    A fire hose held near the ground shoots water at a speed of 6.8m/s. At what angle(s) should the nozzle point in order that the water land 2.0m away? Why are there two different angles?



    2. Relevant equations

    v = v(initial) +at
    x = x(initial) + v(initial)t + .5at^2
    v^2 = v^2(initial) + 2a(x - x(initial)
    x=vt



    3. The attempt at a solution

    I first drew a right triangle and by the problem i was only given the hypotenuse (6.8). Knowing that the adjacent line is the initial velocity in the positive x direction and that the opposite line is the initial velocity in the y positive direction, i tried to solve for theata that way.

    So I did:

    Initial Velocity in the positive X direction is:

    X= X(initial) + V(initial)(t) + .5(a)(t^2)
    2= 0 + (6.8)(cos theta)(t) + 0
    2= (6.8)(cos theata)(t)


    Initial velocity in the positive Y direction is:

    Y= Y(initial) + V(initial)(t) + .5(a)(t^2)
    0= 0 + (6.8)(sin theta)(t) + (-4.9)t^2
    0= (6.8)(sin theta)(t) + (-4.9)t^2

    Once I got to this point I had no idea what to do... any guidance from here would be greatly appreciated. I've never taken Physics before and took geometry around 10 years ago so this has all been erased from my memory..

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



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 11, 2010 #2
    from your first equation, get t in terms of cos theta.
    substitute for t in the second equation.
    you should get a quadratic in tan theta.
     
  4. Jun 11, 2010 #3
    Ok, so if I'm following correctly; solving the first equation for t is:

    t=(2/6.8cos theta)

    Substitute t in equation 1 for that in equation so:

    0 = (6.8sin theta)(2/6.8cos theta) + -4.9(2/6.8 cos theta)2

    How do I make a quadratic and solve for something like this?
     
  5. Jun 12, 2010 #4
    Here have a look very simple method.
    t=2/(6.8cos θ) ...........(1)
    For y-axis
    0=(6.8sinθ)t - 4.9t^2
    t=(6.8sin θ)/4.9............(2)
    Eqn.(1)=Eqn.(2)
    2/(6.8cos θ) =( 6.8sin θ)/4.9
    2sin θ cos θ=4(4.9)/(6.8)^2
    sin2 θ=4(4.9)/(6.8)^2
    θ=........
     
  6. Jun 12, 2010 #5
    Ok I followed you up to here:
    t=2/(6.8cos θ) ...........(1)
    For y-axis
    0=(6.8sinθ)t - 4.9t^2
    t=(6.8sin θ)/4.9............(2)
    Eqn.(1)=Eqn.(2)


    If I set Eqn. 1 = Eqn. 2

    (2/6.8 cosθ) = (6.8 sinθ)t + -4.9t2

    now how do I get rid of the sin and cos, sorry I'm new to this and don't fully understand how to do this.
     
  7. Jun 12, 2010 #6
    You have to see my method. Use 2sinθcosθ =sin 2θ formula.
     
  8. Jun 12, 2010 #7

    rl.bhat

    User Avatar
    Homework Helper

    0 = (6.8sin theta)(2/6.8cos theta) + -4.9(2/6.8 cos theta)2

    Rewrite this equation as

    0 = 2tan(θ) - [4.9*4/(6.8)^2]sec^2(θ)

    0 = 2tan(θ) - [4.9*4/(6.8)^2][1 + tan^2(θ)]

    0 = 2tan(θ) - [4.9*4/(6.8)^2] - [4.9*4/(6.8)^2][tan^2(θ)

    Now solve the quadratic to find two angles.
     
  9. Jun 12, 2010 #8
    I explain simple method.

    t=2/(6.8cos θ) ...........(1)

    t=0 at initial point, t= 2/(6.8cos θ) at final point

    For y-axis
    0=(6.8sinθ)t - 4.9t^2
    t=(6.8sin θ)/4.9............(2)
    Eqn.(1)=Eqn.(2)
    2/(6.8cos θ) =( 6.8sin θ)/4.9
    2sin θ cos θ=4(4.9)/(6.8)^2
    sin2 θ=4(4.9)/(6.8)^2=0.4239
    2θ= 25.81 deg. (or) 180-25.81=154.19 deg.
    θ = 12.54 deg. (or) 77.46 deg.
     
  10. Jun 12, 2010 #9
    Thank you so much Inky, I now get it... I forgot some of the functions and seeing it step by step brought them back to me.

    Thanks again!
     
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