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?