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## Main Question or Discussion Point

Say an object is dropped out of a plane at height h above the ground and flying at a constant speed v0...

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

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!?!?

→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???????????????

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???????????????