We have seen various types of motions in our previous sessions. We also know what happens to objects thrown into space. In this article, we will be discussing projectile **motion** and other concepts like the time of flight formula, horizontal range, maximum height of projectile, the equation of trajectory along with solved examples.

Table of Contents: |

## What is Projectile?

A projectile is any object thrown into space upon which the only acting force is gravity. The primary force acting on a projectile is gravity. This doesn’t necessarily mean that other forces do not act on it, just that their effect is minimal compared to gravity. The path followed by a projectile is known as a trajectory. A baseball batted or thrown is an example of the projectile.

## What is Projectile **Motion**?

When a particle is thrown obliquely near the earth’s surface, it moves along a curved path under constant acceleration that is directed towards the centre of the earth (we assume that the particle remains close to the surface of the earth). The path of such a particle is called a projectile and the **motion** is called **projectile motion**.

**In a Projectile Motion, there are two simultaneous independent rectilinear motions:**

**Along the x-axis:**uniform velocity, responsible for the**horizontal**(forward)of the particle.**motion****Along y-axis:**uniform acceleration, responsible for the**vertical**(downwards)of the particle.**motion**

**Acceleration in the horizontal projectile motion and vertical projectile motion of a particle:** When a particle is projected in the air with some speed, the only force acting on it during its time in the air is the acceleration due to gravity (g). This acceleration acts vertically downward. There is no acceleration in the horizontal direction, which means that the velocity of the particle in the horizontal direction remains constant.

### Parabolic **Motion** of Projectiles

Let us consider a ball projected at an angle θ with respect to the horizontal x-axis with the initial velocity u as shown below:

The **point O** is called the **point of projection**; **θ** is the **angle of projection** and **OB** **= Horizontal Range** or Simply Range. The total time taken by the particle from reaching O to B is called the **time of flight**.

For finding different parameters related to projectile **motion**, we can make use of differential** equations of motions**:

### Total Time of Flight

Resultant displacement (s) = 0 in Vertical direction. Therefore, the time of flight formula is given by using the Equation of **motion**:

gt2 = 2(uyt – sy) **[Here, **u**y**** = u sin θ and **s**y**** = 0]**

**i.e. gt****2**** = 2t × u sin θ**

**Therefore, the**** time of flight formula ****(t) is given by:**

\(Total\,Time\,of\,Flight\,(t)=\frac{2u\sin\Theta}{g}\)

### Horizontal Range

Horizontal Range (OA) = Horizontal component of velocity (ux) × Total Flight Time (t)

R = u cos θ × 2u×sinθg

**Therefore, in a projectile motion the Horizontal Range is given by (R):
**

\(Horizontal\,Range\,(R)=\frac{u^2\sin 2\Theta }{g}\)

### Maximum Height of Projectile

After understanding what is projectile and what is projectile **motion**, let us know the maximum height of projectile. The maximum height of the object is the highest vertical position along its trajectory. The horizontal displacement of the projectile is called the range of the projectile. The range of the projectile is dependent on the initial velocity of the object.

If v is the initial velocity, g = acceleration due to gravity and H = maximum height in meters, θ = angle of the initial velocity from the horizontal plane (radians or degrees).

The maximum height of projectile is given by the formula:

\(H=\frac{v_{0}^{2}sin^{2}\theta}{2g}\)

### The Equation of Trajectory

\(Equation\,of\,Trajectory = x\tan \Theta -\frac{gx^2}{2u^2cos^{2}\Theta }\)

This is the equation of trajectory in projectile **motion**, and it proves that the projectile **motion** is always parabolic in nature.

## Basketball **Physics**

We know that projectile **motion** is a type of two-dimensional **motion** or **motion** in a plane. It is assumed that the only force acting on a projectile (the object experiencing projectile **motion**) is the force due to gravity. But how can we **define** projectile **motion** in the real world? How are the concepts of projectile **motion** applicable to daily life? Let us see some real-life examples of projectile **motion** in two dimensions.

All of us know about basketball. To score a basket, the player jumps a little and throws the ball in the basket. The **motion** of the ball is in the form of a projectile. Hence, it is referred to as projectile **motion**. What advantage does jumping give to their chances of scoring a basket? Now apart from basketballs, if we throw a cricket ball, a stone in a river, a javelin throw, an angry bird, a football or a bullet, all these motions have one thing in common. They all show a projectile **motion**. And that is, the moment they are released, there is only one force acting on them- gravity. It pulls them downwards, thus giving all of them an equal impartial acceleration.

It implies that if something is being thrown in the air, it can easily be predicted how long the projectile will be in the air and at what distance from the initial point it will hit the ground. If the air resistance is neglected, there would be no acceleration in the horizontal direction. This implies that as long as a body is thrown near the surface, the **motion** of the body can be considered as a two-dimensional **motion**, with acceleration only in one direction. But how can it be concluded that a body thrown in the air follows a two-dimensional path? To understand this, let us assume a ball that is rolling as shown below:

**Motion**in one dimension

**motion**? The most common answer would be that it has an x-component and a y-component, it is moving on a plane, so it must be an example of

**motion**in two dimensions. But it is not correct, as it can be noticed that there exists a line which can completely

**define**the

**motion**of the basketball. Thus, it is an example of

**motion**in one dimension. Therefore, the choice of axis does not alter the nature of the

**motion**itself.

**Motion**in Plane

**motion**? It can be seen here that a line cannot

**define**such a

**motion**, but a plane can. Therefore, for a body thrown at any angle, there exists a plane that entirely contains the

**motion**of that body. Thus, it can be concluded that as long as a body is near the surface of the Earth and the air resistance can be neglected, then irrespective of the angle of projection, it will be a two-dimensional

**motion**, no matter how the axes are chosen. If the axes here are rotated in such a way that, then and can completely

**define**the

**motion**of the ball as shown below:

Thus, it can be concluded that the minimum number of coordinates required to completely **define** the **motion** of a body determines the dimension of its **motion**.

## Solved Example For You

### An object is launched at a velocity of 40 m/s in a direction making an angle of 50° upward with the horizontal.

Q1. What is the maximum height reached by the object?

Q2. What is the total flight time (between launch and touching the ground) of the object?

Q3. What is the horizontal range (maximum x above ground) of the object?

**Solution:**

The velocity components \(V_{x}\) and \(V_{y}\) are given by the formula:

In the given problem, \(V_{0} = 40m/s\), ? = 50° and g is \(9.8\,m^{2}\)

The height of the projectile is given by the component y, and it reaches its maximum value when the component \(V_{y}\) is equal to zero. That is when the projectile changes from moving upward to moving downward.

Substituting and solving for t, we get

\(t=\frac{V_{0}\sin(?)}{g} = \frac{40\sin (50^{\circ})}{9.8}=3.12\,seconds\)To find the maximum height, substitute t in the equation y, we get

\(y=\frac{V_0\sin ?}{3.12}-\frac{1}{2}(9.8)(3.12^2)\)Solving, we get

\(y=\frac{40\sin50^{\circ}}{3.12}-\frac{1}{2}(9.8)(3.12^2)=47.9\,meters\)The maximum height reached by the object is 47.9 meters

The time of flight is the interval between when the projectile is launched (t_{1}) and when the projectile touches the ground (t_{2}).

Hence,

\(V_0\sin(?)t-\frac{1}{2}gt^2\)Solve for t

\(t(V_0\sin(?)-(\frac{1}{2}gt)=0\)Solving, we get two solutions as follows:

\(t=t_1=0\) and \(t=t_2=2V_0sin(?)/g\)Time of flight can be calculated as follows:

\(Time\,of\,Flight=2(20)\sin (?)/g=6.25\,seconds\)

Horizontal Range is the horizontal distance given by x at t = t_{2}.

**Suggested Videos for Projectile**

**Motion**### Projectile **Motion** | Kickstart Series

### Half Projectile **Motion**

### Time of Flight in 1D **motion**

### Projectile **Motion** in an Inclined Plane

Hope you learned projectile **motion**, time of flight formula, horizontal range, maximum-height, and the equation of trajectory. Stay tuned with BYJU’S to learn more about projectile **motion** and its applications.

Nice explanation

great notes

helped me a lot!!!!!!!!

great thanks!for your help

wow information. thanks for it

very helpful

Thnku for giving this . It helped me a lot.

thank you very much! It’s very helpful

Thanks so much, helpful it is.

thanks u so much God u

Nice explanation

Thank u

Helped me indeed

helpful

This explanation is super useful,thanks for this.

much helpful very good

helped me so much while doing the notes, thankyou so much byju’s learning.

Wow

Great notes

It is a very nice note

Thank you so much it helped me a lot

It was helpful 😌😌😌😌

👌👌

Very helpful!! Thanks