Horizontal Projection — Trajectory

The situation

A body is projected horizontally with speed $u$ from the origin. Taking rightward as $+x$ and downward as $+y$ (since the body falls downward), what is the shape of the path?

Derivation

The position equations for horizontal projection are:

$x = ut \quad \text{...(horizontal)}$

$y = \frac{1}{2}gt^2 \quad \text{...(vertical, downward positive)}$

From the $x$ equation:

$t = \frac{x}{u}$

Substitute into the $y$ equation:

$y = \frac{1}{2}g\left(\frac{x}{u}\right)^2$

$\boxed{y = \frac{gx^2}{2u^2}}$

Why this is a parabola

The equation $y = \frac{g}{2u^2} x^2$ has the form $y = kx^2$ — a parabola opening in the positive $y$ direction (downward in our convention). There is no linear term in $x$ because there is no $\tan\theta$ term — the launch is horizontal ($\theta = 0$, $\tan 0 = 0$).

Comparing with the general trajectory $y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$, setting $\theta = 0$:

$y = x \cdot 0 - \frac{gx^2}{2u^2 \cdot 1} = -\frac{gx^2}{2u^2}$

The negative sign appears because in the general formula $y$ is measured upward, whereas here we measure $y$ downward. The magnitude is the same.

What determines the shape

$y = \frac{g}{2u^2} \cdot x^2$

The coefficient $\frac{g}{2u^2}$ controls how curved the parabola is:

• Large $u$ → small coefficient → shallow, wide parabola (travels far before falling significantly)
• Small $u$ → large coefficient → steep, narrow parabola (falls quickly)
• Larger $g$ → larger coefficient → steeper parabola

Finding height fallen for a given horizontal distance

$y = \frac{gx^2}{2u^2}$

Example: A ball rolls off a table at $u = 5$ m/s. How far has it fallen when it is $3$ m away horizontally?

$y = \frac{10 \times 9}{2 \times 25} = \frac{90}{50} = 1.8 \text{ m}$