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Ignoring air resistance, the only force on the object is the gravitational force, but it will still move in the direction that the plane was moving, right?

So if

**Fg**is the only force acting on the object,

**Fg**=

**Fnet**=m

**a**

My question is - when solving this equation and integrating, why do you end up with

x(t)= v0*t

y(t)= h - (1/2)gt^2

I'm fully aware that (m/s)(s)= m which explains the x coordinates, but where does v0*t come in when solving the differential equation!?!?

**Fg**= (mgcos270, mgsin270) = (0, -mg)

→mx'' = 0 and my'' = -mg

How can you integrate 0 and get v0*t????????????????

EDIT: I know how to get y(t)= h -(1/2)gt^2

How do you get x(t) = v0*t when mx'' = 0???????????????