# Questions about the downward gravitational force and Newton's second law

## 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 Fg is the only force acting on the object,
Fg=Fnet=ma

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

## Answers and Replies

Related Classical Physics News on Phys.org
integrating one time you get,
dx/dt=c,where c a constant.(integration of zero is a constant and not zero). From the initial condition,c=v0 and integrating one time again you get that.

mfb
Mentor
but it will still move in the direction that the plane was moving, right?
Right, as there is no force in horizontal direction. However, the 3-dimensional "direction of the movement" changes as it gets a component downwards.

but where does v0*t come in when solving the differential equation!?!?
In the first integration of a=0 for the horizontal component, you get an integration constant: v=c1
The second integration then gives x=c1t + c2 (another integration constant). As those integration constants correspond to the initial velocity and initial position, they are called v0 and x0 and you get x(t)=v0t + x0.