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ch3570r

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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 don't 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?

"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 don't 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?

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