- #1

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[tex]y=-\frac{1}{2}gt^2+v_{0}\sin{(\theta)}t+h[/tex]

[tex]x=v_{0}\cos{(\theta)}t[/tex]

Is there any way to use vectors to determine the postion of a projectile? If so, how would you convert rectangular coordinates to vectors?

Thanks.

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- Thread starter amcavoy
- Start date

- #1

- 665

- 0

[tex]y=-\frac{1}{2}gt^2+v_{0}\sin{(\theta)}t+h[/tex]

[tex]x=v_{0}\cos{(\theta)}t[/tex]

Is there any way to use vectors to determine the postion of a projectile? If so, how would you convert rectangular coordinates to vectors?

Thanks.

- #2

James R

Science Advisor

Homework Helper

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[tex]\vec{r} = x(t)\hat{i} + y(t)\hat{j}[/tex]

where the i and j are unit vectors in the x and y directions.

- #3

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What you are asking about is the very base of Newtonian mechanics. It works like this : Suppose we work in two dimensions denoted by a x-axis and an y-axis. You can work in as many dimensions as you want because all you have to do is add a unit vector to the formula's, as you will see.

Starting from the acceleration ,one can calculate the velocity and position by using integrals at any time : r_0 is initial position at t=0, v_0 is initial velocity

[tex]\vec{F} = F_x \vec{e_x} + F_y \vec{e_y} = m(a_x \vec{e_x} + a_y \vec{e_y})[/tex]

Where the e_x and e_y denoted the x and y-direction (ie the unit vectors)

Now, integrating will yield

[tex]\vec{v} = \vec{v_0} + \vec{a}t[/tex]

[tex]\vec{r} = \vec{r_0} + \vec{v_0}t+ \vec{a} \frac{t^2}{2}[/tex]

Now, the trick really is (and that's the essential part) to apply the same procedure in each direction. The procedure i am talking about is projecting each vector along a direction using the triangle-equalities.

For example : in the x-direction you will have :

[tex]F_x = ma_x [/tex]

[tex]v_x = v_{0x} + a_xt[/tex] and

[tex]x = r_{0x} + v_{0x}t+ a_x \frac{t^2}{2}[/tex]

[tex]v_{0x} = ||\vec{v_0}||cos( \theta)[/tex]

[tex]r_{0x} = ||\vec{r_0}||cos( \theta)[/tex]

the ||.|| denotes the MAGNITUDE of the vector

You see ? the clue is that each vector can be written as a sum of an x and y component [tex]\vec{A} = A_x \vec{e_x} + A_y \vec{e_y}[/tex]

[tex]A_x = ||\vec{A}||cos( \theta)[/tex]

[tex]A_y = ||\vec{A}||sin( \theta)[/tex]

You will need to be carefull with the signs of the x and y components because those depend of the direction of each component with respect to the actual x and y axis.

So, when a force is given, like : F = m(2e_x - 9.81e_y) and the initial position has components r_0 = 2e_x + 6e_y and the initial velocity v_0 = 6e_y, can you write down the equations for both velocity and position in each direction ???

In the end , you must realize that this system is very easy because, nomatter how complicated the force may look, the procedure to determin both position and velocity as a function of time is always the same: Projecting the vectors along the given directions. Once, you have done that, you can do almost anything with these formula's...

regards

marlon

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