Why does a cannonball shot from a cannon not descend vertically, but rather curvilinearly? Why does a tossed javelin draw a sumptuous arc before it stings the ground? Regardless of the nature of the projectile, the arc one draws through the air is precisely a parabola.

The reason is, of course, gravity, the only force that affects its motion (neglecting air resistance) after it is projected. However, what we are essentially asking is, why does gravity force it to trace a parabola? Kepler knew that the planets orbited the Sun in an ellipse, but he didn’t know *why* they did so. In that same vein, *Why* doesn’t a projectile trace any other shape but a parabola?

## The Equation

Of course, this is not true when a projectile is projected perpendicular to the Earth’s surface. To observe a parabolic trajectory, we must project it at some angle with the surface. Even though no horizontal forces affect a projectile following its launch, it is the initial horizontal force that makes the glorious journey possible. How else could a javelin travel a horizontal distance if it were not provided with a horizontal force?

In the 17th century, mankind had yet to build a rocket and the most powerful telescopes couldn’t look further than Saturn. Despite these constraints, how could Newton, sequestered in a tiny room in England, discover that the planets orbit the Sun and not – as the most eminent philosophers of the time believed (or rather hoped) – in a circle, but rather an ellipse? Mathematics, of course.

Newton proved Kepler’s claim by discovering a relationship between the distance between Earth and the sun, and the angle it spans while rotating around it. He discovered that it was exactly the same relationship that describes a point tracing an ellipse. However, his evaluation was based on his newly proposed law of gravity. If his law were untrue, his proof would also fall apart. We now know that what he proposed was true; Newton never explained what gravity *is*, but he beautifully explained how it works.

Similarly, to determine which curve a projectile traces, we must find an equation that describes its motion and the curve that corresponds to it.

The projectile is projected with an initial velocity ‘v’ at an angle ‘Φ’ with respect to the surface. The distance the projectile travels horizontally (on the X-axis) is given as *x = vtcosΦ* (v=x/t). However, the distance it travels vertically (on the Y-axis) is given as *y = vtsinΦ – (½)gt²*. This is because vertically, the projectile experiences a force and thus acceleration, namely, the acceleration due to gravity, denoted by ‘g’.

Now, because this acceleration is constant, we can use the kinematic equation *s = ut + (½)at²* to calculate the distance ‘y’. Here, ‘u’ is the initial velocity, which in this case is *vsinΦ* and ‘a’ is the constant acceleration, which in this case is ‘-g’, due to our selected convention. Therefore, the vertical distance *y = vtsinΦ – (½)gt²*.

To find ‘y’ in terms of ‘x’, or to obtain an equation that describes the relationship between ‘y’ and ‘x’, we solve for ‘t’ in the first equation and substitute its value in ‘y’.

or,

Substitute the value of t in:

Here, *tanΦ* and *g/2v²cos²Φ* are constants, so the equation uncannily resembles the equation *y = ax+bx² –* the equation of a parabola!

#### References

**About the Author**:

Akash Peshin is an Electronic Engineer from the University of Mumbai, India and a science writer at ScienceABC. Enamored with science ever since discovering a picture book about Saturn at the age of 7, he believes that what fundamentally fuels this passion is his curiosity and appetite for wonder.

**More from this author**.

These are my favorite type of articles. I forgot how fun it is. I remember doing calculation similar to this but we broke up 2 separate velocities. One vertical velocity and one horizontal velocity. Not including resistance. The vertical is calculated separately giving us time and we can determine where it lands. and the parabola affect is because the vertical instantly starts to slow down until it stops and accelerates once its starts moving back to earth. but the horizontal is consistent.