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# Derivation of Equation of Trajectory of a Projectile Motion | JEE      LIVE
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# 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. The projectile is something like a baseball that is batted or hurled.

A projectile usually follows a curved path, or as it is known in physics, a parabolic trajectory. We all must’ve come across the projectile motion at some point in our lives. On the curved path of the projectile, the acceleration due to gravity is constant, and it acts towards the centre of the Earth. We will now look at the way in which projectile motion can be solved.

## 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{align}&u_{x}=u \cos \theta \\ \\ & u_{y}=u \sin \theta \end{align}

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{align}&v=u-g t \\ \\ & s=u t-\dfrac{1}{2} g t^{2} \\ \\ & v^{2}-u^{2}=2 g s \end{align}

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.

## 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{align}&v_{y}=u_{y}-g t \\ \\ & 0=u \sin \theta-g t \\ \\ & g t=u \sin \theta \\ \\ & t=\dfrac{u \sin \theta}{g} \end{align}

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

$T=\dfrac{2 u \sin \theta}{g}$

## 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,

$v_{x}=\dfrac{s}{t}$

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 vx , which will be the same as ux.

So the range will be,

\begin{align}&R=v_{x} \times T \\ \\ & R=u \cos \theta \times \dfrac{2 u \sin \theta}{g} \\ \\ &R=\dfrac{2 u^{2} \sin \theta \cos \theta}{g} \end{align}

We can further simplify the equation by using,

\begin{align}&\sin 2 \theta=2 \sin \theta \cos \theta \\ \\ & R=\dfrac{u^{2} \sin 2 \theta}{g} \end{align}

## Maximum Height of the Projectile

The 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{align}&v_{y}^{2}=u_{y}^{2}-2 g s \\ \\ & 0=(u \sin \theta)^{2}-2 g H \\ \\ & -u^{2} \sin ^{2} \theta=-2 g H \\ \\ & \dfrac{u^{2} \sin ^{2} \theta}{2 g}=H \end{align}

So the maximum height of a projectile is given as,

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

## 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{align}&v_{y}=u-g t \\ \\ & v_{y}=u \sin \theta-g t \end{align}

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

\begin{align}&y=u t-\dfrac{1}{2} g t^{2} \\ \\ & y=u \sin \theta t-\dfrac{1}{2} g t^{2} \end{align}

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

\begin{align}&u x=\dfrac{x}{t} \\ \\ & u \cos \theta=\dfrac{x}{t} \\ \\ & t=\dfrac{x}{u \cos \theta} \end{align}

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

\begin{align}&y=u \sin \theta \times \dfrac{x}{u \cos \theta}-\dfrac{1}{2}g \times\left(\dfrac{x}{u \cos \theta}\right)^{2} \\ \\ & y=\dfrac{x \sin \theta}{\cos \theta}-\dfrac{g x^{2}}{2u^{2} \cos ^{2} \theta} \\ \\ & y=x \tan \theta-\dfrac{g x^{2}}{2u^{2} \cos ^{2} \theta} \end{align}

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.

## 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 $y=x \tan \theta-\dfrac{g x^{2}}{2 u^{2} \cos ^{2} \theta}$.

Last updated date: 26th Sep 2023
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## FAQs on Derivation of Equation of Trajectory of a Projectile Motion | JEE

1. What are real-life applications of projectile motion?

The projectile motion is used in sports in real life. Real-world instances of projectile motion include playing football and basketball. A basketball player shoots the ball into the basket in such a way that it takes the shape of a parabola throughout its trajectory. The ball creates a curve so that the distance it travels from the fixed point to the other axis is equal to the curve's radius. Real-world applications of projectile motion look like this.

2. What is the weightage of Projectile Motion in JEE Main?

Projectile motion is an extremely important topic in JEE Main. It comes under the chapter of and is basically a two-dimensional kinematic problem. Due to its complexity, the probability of a question being asked about Projectile Motion is high. Incline and oblique projectiles are important from an exam point of view. Usually, one or two questions are asked from kinematics, and its weightage is 3.3%. So prepare this topic well and solve examples related to it.