JEE Main 2025: Derivation of Equation of Trajectory in Physics

The equation of trajectory is a mathematical representation of the parabolic path followed by a projectile under the influence of gravity. This concept is a core of physics, particularly in kinematics, and is frequently tested in competitive JEE Main Examinations. The trajectory equation allows us to determine the position of a projectile at any given point during its motion, linking its horizontal and vertical displacements.

In this detailed guide, we’ll derive the equation of trajectory step by step, using fundamental principles of motion and simplifying assumptions. We’ll also explore how this equation is used to calculate key parameters like the maximum height, time of flight, and horizontal range of a projectile. Understanding this derivation not only prepares you for exam questions but also helps in grasping its applications in fields like ballistics, sports science, and space exploration.

Introduction to Projectile Motion

Before moving on to the equation of trajectory of a projectile motion, it is necessary to first understand the basics of projectile motion. We can define what a projectile means. Any object launched into space with only gravity acting on it is referred to as a projectile. Gravity is the main force affecting a projectile. This doesn't imply that other forces don't affect it; it merely means that their impact is far smaller than that of gravity. A projectile's trajectory is its route after being fired. 

Understanding Projectile Motion:

Imagine throwing a ball into the air. It flies through the air, reaches a peak, and then falls back to earth. This simple act captures the essence of projectile motion, a fundamental concept in physics and a common element in JEE Main questions. But what defines the path of this thrown ball? Can we predict its every point in its journey?

The answer lies in the Equation of Trajectory of Projectile Derivation, a mathematical expression that describes the projectile's position at any given instant. Deriving this equation requires a close examination of the forces at play and a keen understanding of kinematics.

Assumptions and Approach:

Before diving into the derivation, let's establish some simplifying assumptions:

1. We consider the projectile a point mass, neglecting its size and shape.

2. Air resistance is ignored, focusing on the idealized case of gravity as the only external force.

3. The projectile launches at an angle θ with an initial velocity u.

Solution of a Projectile Motion

We can solve the projectile by resolving the motion of the projectile into two independent rectilinear motions along the x and y axes, respectively. 

Projectile motion

Suppose we have a projectile that is projected with an initial velocity u and an angle θ with respect to the x-axis. This angle $\theta$ is known as the angle of the projectile. We can then resolve the velocity u into its x and y components as,

$\begin{array}{rl}& u_x=u\cos \theta \\ \\ & u_y=u\sin \theta \end{array}$

We can then solve the projectile along the x and y axes using these components.

The three equations of motion for a constant acceleration due to gravity will be used to solve the projectile. These equations are,

$\begin{array}{rl}& v=u-gt\\ \\ & s=ut-{\displaystyle \frac{1}{2}}g{t}^2\\ \\ & v^2-u^2=2gs\end{array}$

The acceleration due to gravity is only along the y-direction and so the velocity along the x-axis will remain constant. We can now find important parameters for projectile motion.

Equation of Trajectory of Projectile Motion Derivation at Time of Flight

The time of flight is the total time that the projectile stays in the air, from the moment that it is projected to the moment it hits the ground. We know that the velocity at the highest point is zero. In our case, the highest point is A and at this point, velocity is zero in the y-direction. We can write the equation of motion for this case as,

$\begin{array}{rl}& v_y=u_y-gt\\ \\ & 0=u\sin \theta -gt\\ \\ & gt=u\sin \theta \\ \\ & t={\displaystyle \frac{u\sin \theta }{g}}\end{array}$

This is the time that it takes for the projectile to reach the highest point. 

Now the projectile will again take the same amount of time to fall back to the earth. This means that the time of flight (T) can be written as twice the time that it takes for the projectile to reach the highest point. So the flight time will be

$T={\displaystyle \frac{2u\sin \theta }{g}}$

Equation of Trajectory of Projectile Motion Derivation at Horizontal Range

As the name suggests, horizontal range is simply the distance that the projectile travels in the horizontal direction. In our case, the horizontal range or simply the range is represented by R. We can calculate the range by using the equation of motion in the x-direction. It was already discussed that the velocity along the x-axis remains constant because no acceleration acts in that direction. Velocity can be written as,

$v_x={\displaystyle \frac{s}{t}}$

Here, s is the displacement, and t is the time.

The displacement, in our case, will be the range and the time that the projectile stays in flight will be t. We have already calculated the time of flight, and we know the value of v_x , which will be the same as u_x.

So the range will be,

$\begin{array}{rl}& R=v_x\times T\\ \\ & R=u\cos \theta \times {\displaystyle \frac{2u\sin \theta }{g}}\\ \\ & R={\displaystyle \frac{2u^2\sin \theta \cos \theta }{g}}\end{array}$

We can further simplify the equation by using,

$\begin{array}{rl}& \sin 2\theta =2\sin \theta \cos \theta \\ \\ & R={\displaystyle \frac{u^2\sin 2\theta }{g}}\end{array}$

Equation of Trajectory of Projectile Derivation at Maximum Height 

The Equation Of Trajectory Of Projectile Derivation at maximum height of the projectile can be calculated by using the equation of motion in the y-direction. At the maximum height of the projectile, the velocity in the y-direction will be zero. So we can use this equation to find the maximum height H.

$\begin{array}{rl}& v_y^2=u_y^2-2gs\\ \\ & 0=(u\sin \theta {)}^2-2gH\\ \\ & -u^2{\sin }^2\theta =-2gH\\ \\ & {\displaystyle \frac{u^2{\sin }^2\theta }{2g}}=H\end{array}$

So the maximum height of a projectile is given as,

$H={\displaystyle \frac{u^2{\sin }^2\theta }{2g}}$

Derivation of the Equation of Trajectory

The projectile follows a parabolic trajectory, or as we have said before, a curved path. We have already derived the Time of Flight, Horizontal Range, and Height of the Projectile. But this does not provide the complete solution for a projectile. We need the equation of trajectory for the complete solution because it will provide the relation between the x and y coordinates at any point of time in the motion of the projectile. 

Now we know that the velocity of the projectile in the x-direction is constant throughout the motion. It is,

$u_x=u\cos \theta$

However, in the y-direction, there is the acceleration due to gravity. The velocity in the y-direction changes with time, so the equation for the velocity in the y-direction can be written as,

$\begin{array}{rl}& v_y=u-gt\\ \\ & v_y=u\sin \theta -gt\end{array}$

We can also write the equation for the displacement in the y-direction as,

$\begin{array}{rl}& y=ut-{\displaystyle \frac{1}{2}}g{t}^2\\ \\ & y=u\sin \theta t-{\displaystyle \frac{1}{2}}g{t}^2\end{array}$

Now to relate the x and y directions we can eliminate t from the above equation by using,

$\begin{array}{rl}& ux={\displaystyle \frac{x}{t}}\\ \\ & u\cos \theta ={\displaystyle \frac{x}{t}}\\ \\ & t={\displaystyle \frac{x}{u\cos \theta }}\end{array}$

Substituting this value of t in the equation for y, we get,

$\begin{array}{rl}& y=u\sin \theta \times {\displaystyle \frac{x}{u\cos \theta }}-{\displaystyle \frac{1}{2}}g\times {\left({\displaystyle \frac{x}{u\cos \theta }}\right)}^2\\ \\ & y={\displaystyle \frac{x\sin \theta }{\cos \theta }}-{\displaystyle \frac{g{x}^2}{2u^2{\cos }^2\theta }}\\ \\ & y=x\tan \theta -{\displaystyle \frac{g{x}^2}{2u^2{\cos }^2\theta }}\end{array}$

where,

y - Vertical distance of the object

x - Horizontal distance of the object

u - Initial velocity of the object

θ - The angle of projection with respect to the horizontal

g - Acceleration due to gravity.

This equation is the equation of trajectory for a projectile. It is also known as the equation of the path of a projectile. This was the whole equation of trajectory derivation. If someone asks to derive an equation of the path of the projectile, this whole derivation will be the answer. 

Using the Equation of Trajectory:

The equation of trajectory can be used for various purposes in JEE Main problems:

1. Finding coordinates at any time: Simply plug in the desired time (t) and initial values (u and θ) to calculate the x and y coordinates of the projectile at that instant.

2. Determining maximum height: Differentiate the equation with respect to time and equate to zero to find the time for maximum height. Then, substitute this time back into the equation to find the corresponding maximum height.

3. Calculating horizontal range: Set y to zero and solve for x to find the time at which the projectile hits the ground. This time can then be used to calculate the horizontal range (x cos(θ)). 

The derivation of the equation of trajectory is not just about crunching numbers. It's about understanding the interplay between forces, velocities, and positions. By internalizing this process, you'll gain a deeper appreciation for projectile motion and its applications, giving you an edge in JEE Main and beyond. 

Formulas on Projectile Motion for Key Parameter:

Solved Examples

1. Determine the vertical distance of the ball that is launched from some distance, making the projectile motion. The ball was thrown with an initial velocity of 40 m/s at an angle of 45°, and it hit the ground after 4 seconds. (g = 10 m/s^2)

Solution:

Given: Angle (θ) = 45°

Time = 4 sec

Initial Velocity (u) = 40 m/s

Horizontal distance (x) = $u_{x}t$

x = ucosθ t

x = 40 × cos 45° × 4 = 113.137 m

Using the trajectory formula:

$\begin{array}{rl}y& =x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }\\ y& =113.137\times \tan {45}^{\circ }-\frac{10\times (113.137{)}^2}{2\times (40{)}^2\times {\cos }^2{45}^{\circ }}\\ y& =113.137\times 1-\frac{12799.81}{1600}\\ y& =113.137-79.99=33.147\text{m}\end{array}$

So, the ball covers a vertical distance of 33.147 meters.

2. Determine the vertical distance of the ball that is launched from some distance, making the projectile motion. The ball was thrown with an initial velocity of 13 m/s at an angle of 30°, and it hit the ground after 5 seconds. (g = 9.8 m/s^2)

Solution:

Given: Angle (θ) = 30°

Time = 5 sec

Initial Velocity (u) = 13 m/s

Horizontal distance (x) = $u_{x}t$

$x=u\cos \theta t$

$x=13\times \cos {30}^{\circ }\times 5=56.29\text{m}$

Using the trajectory formula:

$\begin{array}{rl}y& =x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }\\ y& =56.29\times \tan {30}^{\circ }-\frac{9.8\times (56.29{)}^2}{2\times (13{)}^2\times {\cos }^2{30}^{\circ }}\\ y& =56.29\times \frac{1}{\sqrt{3}}-\frac{31,051.928}{253.5}\\ y& =32.49-122.49=-89.99\text{m}\end{array}$

3. A boy threw a stone with a velocity of 9 m/s at an angle of 35°. Find the equation for the path of a projectile using the trajectory formula. (g = 10 m/s^2)

Solution:

Given: Initial Velocity (u) = 9 m/s, Angle (θ) = 35°

Using the trajectory formula:

$\begin{array}{rl}y& =x\tan \theta -\frac{g{x}^2}{2(ucos\theta {)}^2}\\ y& =x\tan {35}^{\circ }-\frac{10\times x^2}{2\times (9\times \cos {35}^{\circ }{)}^2}\\ y& =0.7002x-0.0920x^2\end{array}$

The equation of the projectile's path is $y=0.7002x-0.0920x^2$.

4. A bullet is fired from a gun with a velocity of 15 m/s at an angle of 60°. Find the equation for the path of a projectile using the trajectory formula. (g = 9.8 m/s^2)

Solution:

Given: Initial Velocity (u) = 15 m/s, Angle (θ) = 60°

Using the trajectory formula:

$\begin{array}{rl}y& =x\tan \theta -\frac{g{x}^2}{2(ucos\theta {)}^2}\\ y& =x\tan {60}^{\circ }-\frac{9.8\times x^2}{2\times (15\times \cos {60}^{\circ }{)}^2}\\ y& =x\times \sqrt{3}-\frac{0.087\times x^2}{\sqrt{225}}\end{array}$

The equation of the projectile's path is $y=\sqrt{3}x-0.087x^2$.

Applications of Projectile Motion

• Projectile motion is used in sports like basketball and football to calculate the ideal angle and velocity for accurate shots or long passes. Golfers also rely on it to optimise ball distance and control.

• Military applications include predicting missile and bullet trajectories to hit precise targets. Artillery uses projectile motion to calculate range and angles for effective strikes.

• In space exploration, it helps design rocket launch trajectories and place satellites in stable orbits. Lunar missions also use it to determine landing paths.

• Architects apply projectile motion principles to design parabolic arches and fountains. This ensures stability in structures and visual appeal in water jet displays.

• Video games and animation use projectile motion to simulate realistic paths for objects like arrows or balls. It adds realism to gaming and special effects in movies.

• Everyday examples include throwing stones or using water sprinklers, where understanding projectile motion helps in predicting their path and range.

Conclusion

When a body is launched at a speed that creates an angle with the horizontal, it follows a parabolic trajectory known as a projectile. It can also be defined as any object launched into space with only gravity acting on it. Gravity is the main force affecting a projectile. Other forces are there, but they don’t impact the projectile as much as gravity. The time of flight of a projectile is the time that the projectile stays in flight. The horizontal range is the distance that the projectile covers in the horizontal direction. The time of flight, horizontal range, and maximum height reached by the projectile depends on the initial velocity and the angle of the projectile. The equation of trajectory is given as $y=x\tan \theta -{\displaystyle \frac{g{x}^2}{2u^2{\cos }^2\theta }}$.

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