phy - mechanics

You need to use a 2D graph to have you.

Since the force on horizontal part will not work on the vertical part, you need to seperate them into 2 parts, the horizontal and verticle.

Taing an example, a simple projectile motion.

If I throw a ball mass m with 30 degree with the ground with a velocity 10m/s. (Assume to right hand side)

First I will draw a force diagram, indicating the force and velocity on verticle part and horizontal.

Then you will have
Fv=-mg, Fh=0
vv=10sin(30 degree), vh=10cos(30 degree)

where in verticle part, I take upward as positive. in horizontal I take RHS as positive.

Verticle part
Fv=-mg, vv(t=0)=5m/s
we have a=-g, take g=10m/s^2
s=5t-2.5t^2

For the ball reach the ground, s=0, we have
0=2t-t^2, t=0 or t=2s

Horizntal
Fh=0, vh(t=0)=8.66m/s
t=2, then distance travelled = d = vh(t) x t = 8.66(2)=17.32m.

See, divide into 2 parts to do it.

For an inclined plane, directions are usually taken parallel to the inclined plane. That is, in directions either [upward along the plane] or [downward along the plane].

Although most people would like to take the direction [upward along the plane] as postive, and so [downward along the plane] as negatice, this is not a rule. It is entirely up to your choice in deciding whether you take the [upward along the plane] as positive or negative.