- #1

UrbanXrisis

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A particle of mass m is shot with an initial velocity v, and makes and angle [tex]\theta[/tex] with the horizontal as show http://home.earthlink.net/~urban-xrisis/clip001.jpg [Broken]

The particle moves in the gravitational field of the Earth. What is the momentum of the particle at the highest point of its trajectory?

since I honestly do not have a clue as to how to do this problem, I will say what I know.

[tex]angular momentum=L=rXp=rXmv=rmvsin\theta[/tex]

[tex]r=\frac{R}{2}tan\theta[/tex]

[tex]L=rmvsin\theta[/tex]

[tex]L=\frac{R}{2}tan\theta mv sin\theta[/tex]

[tex]L=\frac{rmvsin^2\theta}{2cos\theta}[/tex]

the actual answer is:

[tex]L=\frac{-mv^3sin^2\theta cos\theta}{2g}k[/tex]

it seems like I have to use the cross product to get that answer. I'm acually really lost.

The particle moves in the gravitational field of the Earth. What is the momentum of the particle at the highest point of its trajectory?

since I honestly do not have a clue as to how to do this problem, I will say what I know.

[tex]angular momentum=L=rXp=rXmv=rmvsin\theta[/tex]

[tex]r=\frac{R}{2}tan\theta[/tex]

[tex]L=rmvsin\theta[/tex]

[tex]L=\frac{R}{2}tan\theta mv sin\theta[/tex]

[tex]L=\frac{rmvsin^2\theta}{2cos\theta}[/tex]

the actual answer is:

[tex]L=\frac{-mv^3sin^2\theta cos\theta}{2g}k[/tex]

it seems like I have to use the cross product to get that answer. I'm acually really lost.

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