Paramet. eqn. of traject. w/ s&w gravity

[shoopa]

hello all.
Im a bit stuck on a math problem. I am trying to figure out what the parametric equations of a trajectory with a gravitational force pulling west and south. My first questions are should the t variable be squared and the gravity multiplied by .5 in:
x=(v(o)cos(angle))t-.5gt2
I think this because its counterpart, y, has this.

would it also go something like:
distance=x= [((v(0)^2)sin(2*angle)]/ [g(w)*g(s)]
where g(w) represents gravity pulling west, and g(s) for gravity south. What i question in this is is g(w) is multiplied by g(s). i think this is correct but i would really appreciate a confirmation from a math wiz (the people on this site). Thanks!

 

[HallsofIvy]

Is there a reason for doing such a thing? Yes, you can set up a coordinate system so that gravity is pulling "west and south", that is, at an angle to you NS, EW coordinate axes, but it makes things a lot harder! In fact that's why we HAVE "up" and "down". It's so much easier to have gravity pulling in the direction of one axis that we tend to THINK that way!

 

[M98Ranger]

Hi there,
I'm not a math wiz, but I can try to help with your question. The parametric equations for a trajectory with a gravitational force pulling west and south would be:
x = (v0 cosθ)t
y = (v0 sinθ)t - 0.5gt^2

Where v0 is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity (which should be negative in this case since it is pulling south). The t variable should not be squared in the x equation, as it is only used to represent time and not acceleration. And the 0.5 in the y equation is necessary to account for the acceleration due to gravity.

As for your second question, the distance formula you provided is incorrect. The correct formula for distance is:
distance = (v0^2 sin2θ)/g

There is no need to multiply g(w) and g(s) together, as they are both components of the same gravitational force. I hope this helps clarify things for you. Good luck with your math problem!