# Parabolic Path of a Projectile

1. Jul 11, 2009

### DJ24

How do we know, or how can we prove that the path of a projectile is (at least part of) a parabola?

2. Jul 11, 2009

### nicksauce

Easily:

In the vertical direction
$$F = ma = m\frac{d^2 y}{dt^2} = -mg$$

Integrated twice we obtain
$$y(t) = y(0) + v_y(0)t - \frac{1}{2}gt^2$$

Which is indeed the equation of a parabola with respect to time.

Now in the x-direction, we have
$$x(t) = x(0) + v_x(0)t$$
since there is no force in this direction. It should be clear that if we re-write y(t) in terms of x(t), by solving for t in the above equation, y(t) will also be a parabola in terms of x.

3. Jul 11, 2009

### rcgldr

A parabola occurs under the simplified case of a flat earth and a constant gravitational field and no aerodynamic drag. For a spherical earth, with no drag, and the equivalent of a point source for gravity, the path is part of an ellipse.