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Need Help w/ Projectile Motion Problem

  1. Nov 10, 2006 #1
    I've spent the past two hours or so working on some homework problems, and now I'm stuck on one question:

    "When a water gun is fired while being held horizontally at a height of 1.00 m above ground level, the water travels a horizontal distance of 5.00 m. A child, who is holding the same gun in a horizontal position, is also sliding down at a 45.0degree incline at a constant speed of 2.00 m/s. If the child fires the gun when it is 1.00 m above the ground and the water takes 0.329 s to reach the ground, how far will the water travel horizontally?"

    The answer is 4.11m, but I dont know how to get that.

    My knowns:

    Y = 1
    X = 5
    45 degrees
    Vx = 2m/s
    t = .329(s)
    Change in X = ??

    Equations (▲ = change in : Ø = degree: G = Gravity 9.8m/s)
    ▲Y = Vi*sinØ-G*▲t
    Vy = Vi*sinØ

    ▲x = Vi*cosØ-▲t
    Vx = Vi*cosØ
    Vi^2 = (g*▲x)/(2*sinØ*cosØ)

    Anyone have any ideas?
     
    Last edited: Nov 10, 2006
  2. jcsd
  3. Nov 10, 2006 #2
    Use what they tell you about the normal firing (squirting, really) of the gun to find the initial velocity given to the water by the gun.

    Then notice that the child, and hence the gun, are moving with a velocity of 2m/s not in the x direction, but at a 45 degree angle.

    So when the kid squirts the gun, the water gets an initial velocity both from the squirting mechanism of the gun and the sliding that's going on.

    Does that help?
     
  4. Nov 10, 2006 #3
    hopefully it helps, I'll try ur advice later (im busy at the moment)

    thanks
     
  5. Nov 10, 2006 #4
    well, that did help w/ better understanding the problem, but I still cant get 4.11m as an answer. It makes sense that it travels less, but im having trouble w/ what equation(s) to use. Would the initial velocity be 2, because its a constant speed....or is it something else??
     
  6. Nov 11, 2006 #5

    radou

    User Avatar
    Homework Helper

    As stated before, the initial velocity consists of two contributions: the initial velocity of the gun and the initial velocity of the sliding. So, the initial velocity in the x direction is Vox = Vox(sliding) + Vox(gun) = 2.0 * cos(45) + Vox(gun). You can calculate the initial velocity of the gun easily, since the height and horizontal distance are given.

    So, the only equation you need now is d = Vox * t, where the time t is given.
     
  7. Nov 11, 2006 #6
    thanks radou, that really helped me out
     
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