Hi Emily.

How do you prove it is a parabola? Well, it depends on what you start with. If you can take as fact that the basic equations of projectile motion in physics are correct, then you can prove it algebraically.

A projectile (such as a drop of water) ejected from the fountain has an initial velocity V₀ which has components V_xo and V_y0 which are horizontal and vertical components of the initial velocity. If you
ignore air resistance (which you should or you'll never get the parabola you are hoping for), then V_x0 is actually the horizontal velocity throughout the flight of the drop. So we'll just call it V_x .

However V_y does change. It starts at V_y0 upwards (let's call upwards positive) and when the drop reaches the top of the arc, it has reduced to 0. Then it increases in magnitude, but becomes negative, until it
strikes the pool of water at the same speed downwards as it started with upwards. I'm assuming here that the jet of water is at the same elevation as the pool into which it falls.

The equation of projectile motion that describes the vertical distance (height) of the drop is just

y = V_y0 t + (1/2)gt²,

where t is the time in seconds and g is the gravitational acceleration (negative) in distance per second.

The equation of the projectile's horizontal motion is unaccelerated, so it is just x = V_x t.

Notice that I am calling the starting point where the drop is ejected from the jet of water is the origin (y₀ = 0 and x₀ = 0).

Now we have these two equations that describe what x and y are for a given initial velocity with respect to time. This is what we call a parameterized graph.

So let's see if this is a parabola. If it is, we should be able to write it without reference to t, but with known quantities involved in the form y = ax² + bx + c, where a, b and c are just scalar quantities based on known initial conditions (ie. the initial velocity components).

First, if x = V_x t, then t = x/V_x

So since y = V_y0 t + (1/2)gt², we can substitute for t with the expression above.

y = V_y0 (x/V_x ) + (1/2)g(x/V_x )²

which we can reorganize as:

y = [ g / (2 V_x ²) ] x² + [ V_y0 / V_x ] x + [ 0 ]

Since this is indeed in the form of a parabola, this shows that the fountain arc is indeed parabolic.

Emily, if you aren't allowed to use the basic physics equations of motion as starting points for your proof, you will have to try another approach. Experimentally, you could take a digital camera picture of a real fountain and then superimpose a graph of an appropriate parabola to show that it is "experimentally consistent" with a parabolic arc. But choose a calm day so that wind resistance doesn't affect your measurements!

Stephen La Rocque.>