KAHIBARO
Table of Contents
Shape of a Projectile's Path
The trajectory equation gives the path of a projectile as a direct relationship between its vertical position and its horizontal position. Instead of describing motion with time, it answers a different question, where is the projectile vertically when it has moved a certain horizontal distance?
For projectile motion without air resistance, the path is a parabola. This comes from the fact that horizontal motion is uniform, while vertical motion has constant acceleration due to gravity.
Starting Point of the Derivation
For an obliquely launched projectile, let the initial position be $(x_0, y_0)$, the launch speed be $v_0$, and the launch angle be $\theta$ above the horizontal.
The horizontal and vertical positions as functions of time are
$$
x = x_0 + v_0 \cos\theta \, t
$$
$$
y = y_0 + v_0 \sin\theta \, t - \frac{1}{2}gt^2
$$
To get the trajectory equation, eliminate the time variable $t$.
From the horizontal equation,
$$
t = \frac{x - x_0}{v_0\cos\theta}
$$
Substitute this into the vertical equation:
$$
y = y_0 + v_0\sin\theta \left(\frac{x - x_0}{v_0\cos\theta}\right) - \frac{1}{2}g\left(\frac{x - x_0}{v_0\cos\theta}\right)^2
$$
Simplifying gives
$$
y = y_0 + (x - x_0)\tan\theta - \frac{g(x - x_0)^2}{2v_0^2\cos^2\theta}
$$
This is the trajectory equation.
For projectile motion launched with speed $v_0$ at angle $\theta$ from $(x_0, y_0)$, the trajectory is
$$
y = y_0 + (x - x_0)\tan\theta - \frac{g(x - x_0)^2}{2v_0^2\cos^2\theta}
$$
This equation is valid only when air resistance is neglected and gravitational acceleration $g$ is constant.
Meaning of the Terms
The equation has three parts. The term $y_0$ gives the starting height. The term $(x - x_0)\tan\theta$ is the straight line the object would follow if gravity did not act. The last term,
$$
\frac{g(x - x_0)^2}{2v_0^2\cos^2\theta}
$$
bends the path downward, and this is the effect of gravity.
Because the $x^2$ term appears with a negative sign, the curve opens downward.
Special Case, Launch from the Origin
If the projectile starts from the origin, so that $x_0 = 0$ and $y_0 = 0$, the equation becomes simpler:
$$
y = x\tan\theta - \frac{gx^2}{2v_0^2\cos^2\theta}
$$
This is the form most often used in introductory problems.
If the launch point is the origin, the trajectory equation is
$$
y = x\tan\theta - \frac{gx^2}{2v_0^2\cos^2\theta}
$$
This shows clearly that the trajectory is a parabola in $x$ and $y$.
Why the Path is Parabolic
A parabola appears whenever one variable depends linearly on time and another depends quadratically on time. In projectile motion, horizontal position changes as $t$, while vertical position changes as $t^2$ because of gravitational acceleration. Eliminating $t$ leaves a quadratic equation in $x$.
This is a mathematical signature of constant downward acceleration.
Interpreting the Graph
If you graph $y$ versus $x$, the projectile starts at $(x_0, y_0)$, rises if the launch has an upward component, reaches a highest point, and then falls. The slope of the curve changes continuously. At first it is positive, at the top it is zero, and after that it becomes negative.
The initial slope is
$$
\tan\theta
$$
so the launch angle determines the initial direction of the path.
Comparison with Straight Line Motion
Without gravity, the object would move in a straight line:
$$
y = y_0 + (x - x_0)\tan\theta
$$
Gravity adds the negative quadratic correction. This comparison helps show that the trajectory equation is a straight line minus a downward-curving term.
| Situation | Equation for path |
|---|---|
| No gravity | $y = y_0 + (x - x_0)\tan\theta$ |
| With gravity | $y = y_0 + (x - x_0)\tan\theta - \dfrac{g(x - x_0)^2}{2v_0^2\cos^2\theta}$ |
Using the Trajectory Equation
The trajectory equation is useful when time is not needed directly. For example, if you want to know whether a projectile clears a wall located at horizontal position $x$, you can substitute that value of $x$ into the equation and compute $y$.
It is also useful for comparing different launch angles and speeds, since the effect of each appears directly in the formula.
A larger $v_0$ makes the downward curvature smaller, because the denominator becomes larger. A larger $\theta$ increases the initial slope, because $\tan\theta$ becomes larger.
Important Limits and Conditions
The equation works for ideal projectile motion only. It assumes a constant gravitational field and no air drag. Real projectiles such as balls, arrows, or rockets may deviate from a perfect parabola if air resistance is important.
Also, the formula requires $v_0\cos\theta \neq 0$. That means it does not apply to a purely vertical launch, because then horizontal motion is absent and $x$ cannot be used to eliminate time in this way.
The trajectory equation in terms of $y(x)$ requires horizontal motion, so it is not suitable for a purely vertical launch where $\cos\theta = 0$.
Final View
The trajectory equation is the mathematical description of the projectile's path in space. By removing time from the motion equations, it shows directly that projectile motion follows a parabola under constant gravity. It is one of the clearest examples of how horizontal and vertical motions combine into a single curved path.