

Stepwise Guide to Deriving the Trajectory Equation with Examples
The derivation of the equation of trajectory is a fundamental part of projectile motion in physics. This equation describes the path that a projectile follows under the action of gravity, providing the mathematical foundation for analyzing and predicting the motion of objects projected into the air.
Definition of Trajectory and Key Variables
The trajectory of a projectile refers to the curved path followed by an object under the influence of uniform gravity, neglecting air resistance. The standard derivation uses variables such as initial velocity ($u$), angle of projection ($\theta$), and gravitational acceleration ($g$).
| Symbol | Physical Quantity |
|---|---|
| $u$ | Initial velocity (m/s) |
| $\theta$ | Angle of projection (radian) |
| $g$ | Acceleration due to gravity (m/s$^2$) |
| $x$ | Horizontal displacement (m) |
| $y$ | Vertical displacement (m) |
A clear understanding of each variable is essential for solving projectile motion problems and avoids common confusion regarding direction and unit usage.
Stepwise Derivation of the Equation of Trajectory
Consider a particle projected with initial velocity $u$ at an angle $\theta$ with respect to the horizontal axis. Gravity acts downwards with acceleration $g$. The motion can be resolved into two components: horizontal (x-axis) and vertical (y-axis).
The horizontal velocity component is $u\cos\theta$, which remains constant throughout the flight because no horizontal force acts (ignoring air resistance). The corresponding displacement after time $t$ is:
$x = u\cos\theta \cdot t$
The vertical velocity component $u\sin\theta$ is affected by gravity. The vertical displacement after time $t$ is:
$y = u\sin\theta \cdot t - \dfrac{1}{2}gt^2$
Eliminate $t$ from these equations to express $y$ as a function of $x$:
$t = \dfrac{x}{u\cos\theta}$
Substitute this value of $t$ into the equation for $y$:
$y = u\sin\theta \left(\dfrac{x}{u\cos\theta}\right) - \dfrac{1}{2}g\left(\dfrac{x}{u\cos\theta}\right)^2$
$y = x\tan\theta - \dfrac{g x^2}{2u^2 \cos^2\theta}$
This is the general equation of the trajectory for a projectile launched from the origin over a horizontal surface. It mathematically represents the path traced during the motion.
For further details on the vector breakdown of motion components, refer to the Kinematics Overview.
Parabolic Character of the Trajectory Equation
The derived equation $y = x\tan\theta - \dfrac{g x^2}{2u^2 \cos^2\theta}$ contains a linear term ($x\tan\theta$) and a quadratic term in $x$ ($-\dfrac{g x^2}{2u^2 \cos^2\theta}$). The absence of a $y^2$ term confirms that the path is a parabola.
A trajectory is classified as parabolic if it is described by an equation where one variable is of degree two and the other of degree one. Here, the $x^2$ term ensures the path is always a parabola if gravity is constant.
The parabolic nature holds true in the absence of air resistance and for constant $g$. For more on projectile types, see Projectile Motion Explained.
Special Cases in the Equation of Trajectory
Certain particular cases simplify the trajectory equation. For horizontal projection ($\theta = 0^\circ$), the equation reduces to $y = -\dfrac{g x^2}{2u^2}$ since $tan\,0^\circ = 0$ and $cos\,0^\circ = 1$.
For vertical projection ($\theta = 90^\circ$), $cos\,90^\circ = 0$, so the range is zero and the projectile rises and falls along a straight line.
If projected from a height $h$, the formula modifies to $y = h + x\tan\theta - \dfrac{g x^2}{2u^2 \cos^2\theta}$. This adjusted equation frequently appears in advanced problems and examinations.
Summary of Steps for JEE Main and Class 11 Examinations
- Resolve initial velocity into $u\cos\theta$ and $u\sin\theta$
- Write horizontal and vertical displacement equations
- Eliminate $t$ to relate $x$ and $y$ directly
- Simplify the resulting formula to standard parabolic form
- Always use SI units in calculations
Accurate computation of each step is necessary when solving projectile motion numericals. Always verify angle units (radian for calculations, degrees for diagrams) and sign convention for gravity (downwards negative $y$ direction).
For deeper conceptual understanding, students may consult Motion In One Dimension, focusing on kinematic equations derivation.
Applications and Relevance of the Trajectory Equation
The derived trajectory equation is used extensively in JEE Main, NEET, and board-level examinations, helping to calculate projectile height, range, time of flight, or spatial coordinates.
Real-world applications include sports ballistics, engineering, forensics, and any system involving objects projected under gravity. The parabolic path is also essential for graphical analysis of motion and theoretical calculations.
Practicing trajectory-based numericals strengthens understanding and problem-solving skills, which are core requirements for JEE aspirants. Related concepts can be explored in Velocity Of Object And Image.
Key Points for Exam Preparation
- Memorize the standard trajectory equation
- Know unit conversions and standard sign conventions
- Practice by solving for angle, range, or coordinate using provided values
- Identify the nature of the curve as a parabola for constant $g$
- Recognize formula modifications for horizontal or elevated projections
Understanding the mathematics behind trajectory curves enables quick option elimination in competitive exams. For equations involving mass or force, refer to De Broglie Equation as an example of formula application.
FAQs on Derivation of Equation of Trajectory Explained for Students
1. What is the equation of trajectory of a projectile?
The equation of trajectory describes the curved path followed by a projectile under the influence of gravity assuming no air resistance. It is typically represented as:
• y = x tan θ – (gx2)/(2u2cos2θ)
• Here, u is the initial velocity, θ is the angle of projection, g is acceleration due to gravity, x and y are horizontal and vertical coordinates, respectively.
This equation is key to understanding projectile motion in physics for board exams like CBSE Class 11.
2. How is the equation of trajectory derived for a projectile?
The equation of trajectory is derived using the equations of motion along x and y axes:
• Write displacement equations: x = u cosθ ⋅ t and y = u sinθ ⋅ t – (1/2) g t2.
• Eliminate t by substituting t = x / (u cosθ) into the equation for y.
• Simplify to get: y = x tanθ – (g x2)/(2u2 cos2θ).
This formula portrays the parabolic path followed by a projectile.
3. Why is the path of a projectile parabolic?
The path of a projectile is parabolic because the vertical motion is influenced by gravity while the horizontal velocity remains constant:
• Vertical acceleration due to gravity (g) makes the motion downward.
• The horizontal component has no acceleration (if air resistance is neglected).
• Combining these motions using kinematic equations results in a quadratic relationship (parabola) between x and y.
This principle is important in Class 11 Physics and competitive exams.
4. What assumptions are made during the derivation of trajectory equation?
The derivation of the trajectory equation assumes:
• Air resistance is negligible.
• Acceleration due to gravity (g) is constant and acts downward.
• The Earth's surface is flat for the range of motion.
These simplifying assumptions help in obtaining accurate mathematical models suitable for school syllabus contexts.
5. What is the significance of the angle of projection in a projectile's trajectory?
The angle of projection (θ) determines the initial direction and shape of the projectile's path:
• Maximum range occurs at 45° on level ground.
• For a given speed, different angles yield different ranges and heights.
• The angle directly affects the coefficients in the trajectory equation.
Understanding this helps solve numerical problems in physics exams.
6. Explain horizontal and vertical components of projectile motion.
In projectile motion:
• Horizontal component (u cosθ): Constant velocity, no acceleration.
• Vertical component (u sinθ): Changes due to acceleration by gravity (g).
• These two components act independently but together define the parabolic trajectory as per the equation.
This split is essential for problem-solving and understanding motion in two dimensions.
7. Derive the equation of trajectory of a projectile launched with initial velocity u at angle θ.
To derive the equation of trajectory:
• Let initial velocity be u and angle of projection be θ.
• Time to reach point (x, y): t = x / (u cosθ).
• Vertical position: y = u sinθ ⋅ t – (1/2)g t2.
• Substitute value of t into the above:
y = x tanθ – (g x2)/(2u2 cos2θ).
This shows how projectile motion forms a parabola.
8. What factors affect the range of a projectile?
The range of a projectile depends on:
• Initial velocity (u)
• Angle of projection (θ)
• Acceleration due to gravity (g)
• The relation: Range R = (u2 sin2θ)/g
Optimising these factors determines the distance travelled by the projectile.
9. What is the physical meaning of the term 'trajectory'?
Trajectory refers to the path that a projectile follows through space:
• In physics, it represents the locus of positions of the moving object.
• Mathematically, it is described by a parabola for projectiles under gravity.
This term is commonly used in mechanics and is vital for exam answers.
10. What are the main applications of the equation of trajectory?
The equation of trajectory is widely used in:
• Sports (predicting ball paths)
• Military science (projectile launches)
• Engineering (design of fireworks, fountains)
• Physics education (CBSE, NEET, JEE)
These applications utilize the parabolic motion equation to solve real-world problems and exam-based questions.































