Projectile Motion can be seen in our daily life very easily as from throwing a rock to launching a cannonball are all examples of Projectile Motion. It is one of the fascinating topics in the field of physics which has very wide real-world applications. From sports to military technologies all leverage the understanding of Projectiles and their motion under the force of gravity. Understanding Projectile motion helps us predict the trajectory, velocity, and range of objects that are thrown, launched, or dropped in the air. In this article, we will learn the key concepts and formulas of projectile motion and use those to solve real-world scenario-based problems.

## What is Projectile?

The object which is thrown at some angle with some initial velocity performs the parabolic motion under the influence of the force of gravity and is called the projectile. The path of the projectile is known as the trajectory of the motion. All the objects thrown into the air such as stones by some random child, football kicked by some athlete, cricket balls thrown by the player or hit by the batsman, etc. are all examples of projectiles.

**Projectile Motion Definition**

**Projectile Motion Definition**

A projectile is any object thrown into space with only gravity acting on it. The primary force acting on a projectile is gravity. This is not to say that other forces do not act on it; rather, their impact is minimal when compared to gravity.

When a particle is thrown obliquely near the earth’s surface, it follows a curved path with constant acceleration toward the earth’s center (we assume that the particle remains close to the surface of the earth). The path of such a particle is known as a

, and its motion is known asProjectile Path.Projectile Motion

### Properties of Projectile Motion

Projectile motion is one of the most common types of motion in a plane. The only acceleration acting in a projectile motion is the vertical acceleration caused by gravity (g). As a result, equations of motion can be used separately in the X- and Y-axes to determine the unknown parameters.

- The motion of a projectile in two dimensions is divided into two parts:

- Horizontal motion in the x-direction with no acceleration and
- Vertical motion in the y-direction with constant acceleration due to gravity.
- To simplify calculations, projectile motion is typically calculated
.without accounting for air resistance

## Example of Projectile Motion

Projectile motion is an example of Projectile Motion, as it occurs in a two-dimensional plane. The motion of a projectile i.e., an object performing the projectile motion, is under the influence of gravity due to which in the absence of gravity i.e., space, the objects do not perform the projectile motion. Let’s consider examples of projectile motion in the real world.

One familiar example is a ball thrown in the air by some player. When a player throws a ball in the air, the ball follows a projectile motion. The momentum of the player gives the player an advantage by allowing them to adjust the trajectory of the ball. Besides this, other examples of projectile motion include throwing a basketball, a stone into a river, a javelin, an angry bird, kicking a football, or firing a bullet.

All the above-mentioned motions share the common characteristic of being influenced primarily by gravity after release. This means that when an object is thrown into the air, we can predict its duration in the air and the distance it will travel before hitting the ground, assuming we neglect air resistance. In the horizontal direction, there is no acceleration if air resistance is ignored. Consequently, when an object is thrown near the Earth’s surface, its motion can be treated as two-dimensional, with acceleration occurring in one direction.

To illustrate this point, let’s consider a rolling ball. If the ball follows a specific path, it may seem like it is moving in two dimensions. However, if the ball rolls in a straight line, it becomes the motion in one dimension. The choice of axis does not change the nature of the motion itself.

On the other hand, when a ball is thrown at an angle with some initial velocity, its velocity has components in the x, y, and z directions. It might seem like this motion is three-dimensional. However, a plane can fully describe this type of motion, indicating that it is a two-dimensional motion. Regardless of the projection angle, as long as the body remains close to the Earth’s surface and air resistance is negligible, the motion will be two-dimensional. The choice of axes can be rotated to define the ball’s motion entirely.

In conclusion, the dimension of an object’s motion is determined by the minimum number of coordinates needed to fully describe its motion. Projectile motion occurs in a two-dimensional plane and is applicable to various real-life scenarios.

## Terms Related to Projectile Motion

Consider the following example of a ball that is projected at an angle θ from the point O with respect to the horizontal x-axis with an initial velocity u:

Before understanding the derivation of the relation for projectile motion let’s first introduce some terms used in it, which are:

**Angle of Projection**

**Angle of Projection**

The angle at which the body is projected with respect to the horizontal is referred to as the angle of projection. In the above diagram, θ is the angle of projection

**Velocity of Projection**

**Velocity of Projection**

The velocity with which the body is thrown is referred to as the velocity of projection. Here, u is the initial velocity of the projectile that has vertical and horizontal components as shown in the diagram.

**Point of Projection**

**Point of Projection**

A point of projection is the point from which the body is projected in the air. In the above diagram, point O is known as the point of projection.

**Projectile Trajectory**

**Projectile Trajectory**

The path taken by a projectile in the air is referred to as the projectile’s trajectory and in the diagram, the path followed by the projectile is the trajectory of a projectile.

**Horizontal Range**

**Horizontal Range**

The horizontal distance travelled by the body performing projectile motion is referred to as the range of the projectile, and in the above-mentioned diagram OB is Horizontal Range.

## Equation of Motion for Projectile

We know that the linear equations of motion are:

v = u + at

S = ut + 1/2(at^{2})

v^{2}= u^{2}+ 2aS

Applying the above equation for projectile motion the equation will be:

v = u – gt

S = ut – 1/2(gt^{2})

v^{2}= u^{2}– 2gSWhere,

is Initial Velocity,uis Final Velocity,vis Acceleration due to gravity,gis Displacement, andSis Time.t

## Projectile Motion Formula

There are various formulas for the Projectile Motion for calculation of various this such as:

- Time of Flight
- Horizontal Range
- Maximum Height

Let’s discuss these formulas with various different cases as follows:

**Time of Flight of Projectile Motion**

**Time of Flight of Projectile Motion**

Time of flight is the total time taken by the projectile from start to end. We can calculate it as,

In the Y direction total displacement (S_{y}) = 0

Taking motion in Y direction only,

S_{y }= u_{y}t – 1/2(gt^{2})For object to achive peak height u

_{y}= u sinθ and S_{y }= 0, and t it the time taken by object to achive the peak height.0 = usinθ – 1/2(gt

^{2})⇒ t = 2usinθ/g

Time of Flight (2t) = 2usinθ/g

Now there can be various cases of the above-mentioned formula, let’s consider the following cases:

** Case 1: I**f θ = 90°

As we can see from the formula of Time of flight, time taken by the projectile is directly proportional to the angle of projection. For any given initial velocity(u) will be constant and g is always constant i.e., g=-9.8 m/s

^{2}.When projectile is projected at an angle of 90° time of flight will be maximum.

t_{max}= 2usinθ/g = 2u/g[As sin 90° = 1]

** Case 2:** If θ = 30°

When the projectile is projected at an angle of 30° time of flight is half of the t

_{max}as sin30° = 1/2.

t = 2usin30°/g = t_{max}/2

**Horizontal Range of Projectile**

**Horizontal Range of Projectile**

The horizontal range is the distance covered by the projectile horizontally and it can be calculated by the distance = speed/time formula, where speed is the horizontal component of initial speed or velocity and time is the total time of flight. Thus, the formula for Horizontal Range is given by:

**Range (R) = u**_{x}** × t**

And as u_{x} = u cosθ and t = 2usinθ/g

**Range (R) = ucosθ × 2usinθ/g**

As a result, the Horizontal Range of the projectile is given by (R):

Horizontal Range (R) = u^{2}sin2θ/g

Now there can be various cases of the above-mentioned formula, let’s consider the following cases:

** Case 1: I**f θ = 90°

When projectile is projected at an angle of 90° Horizontal range will be zero, because projectile will strike at the same point where the projectile is projected.

R = u^{2}sin2θ/g = 0[As sin 2θ = sin 180 = 0, at θ = 90°]

** Case 2: I**f θ = 45°

When projectile is projected at 45° Horizontal Range of the projectile is maximum.

R_{max}= u^{2}sin2θ/g = u^{2}/gAs sin 90 = 1 and it is the maximum value of the trigonometric ratio sin.

**Maximum Height of Projectile **

**Maximum Height of Projectile**

It is the highest point of the particle (point A). When the ball reaches point A, the vertical component of the velocity (V_{y}) will be zero.

0 = (usinθ)^{2 }– 2gH_{max }

_{ }[ Here,S = H,_{max}v_{y}and= 0u_{y}]= u sin θTherefore, the Maximum Height of the projectile is given by (H

_{max}):

Maximum Height (H_{max}) = u^{2}sin^{2}θ/2g

Now there can be various cases of the above-mentioned formula, let’s consider the following cases:

** Case 1:** if θ = 90°

If we project a projectile at an angle of 90° it achieves maximum height (H

_{max}).

H_{max}= u^{2}sin^{2}θ/2g = u^{2}/2g[As, sin

^{2 }90° = 1 ]

** Case 2:** if θ = 45°

When the projectile is projected at an angle of 45°, the height of the projectile is half of its maximum height (Hmax) as sin

^{2}45° = 1/2.

H = u^{2}sin^{2}θ/2g = (1/2)u^{2}/2g = H_{max}/2We can also say that if the projectile angle is 45° than Horizontal range of projectile will be 4 time the height of projectile.

H = u^{2}/4g = R/4OR

R = 4H[ As Horizontal range at θ = 45°, R = u

^{2}/g ]

**Equation of Trajectory of Projectile**

**Equation of Trajectory of Projectile**

The equation of the trajectory is a path followed by the particle during the projectile motion. The equation is:

y = x tanθ – gx^{2}/2u^{2}cos^{2}θ

**Derivation of Equation of Trajectory of Projectile**

**Derivation of Equation of Trajectory of Projectile**

Let’s consider a projectile launched at an angle θ to the horizontal with an initial velocity u_{o}. Assuming there is no air resistance, the only force acting on the projectile is the force of gravity, which acts vertically downward.

The equation of motion in the x direction is given by:

x = u_{o }cos θ × t . . .(i)

Where,

is the horizontal distance travelled by the projectile,**x**is the time elapsed, and**t****u**is the initial velocity, and θ is the angle of launch._{o}

The equation of motion in the y direction is given by:

y = u_{o}sin(θ) × t – 1/2 × g × t^{2}. . .(ii)

Where,

is the vertical distance travelled by the projectile,**y**is the acceleration due to gravity (approximately 9.8 m/s^2), and**g**is the time elapsed.**t**

From equation (i), **t = x/(u**_{o }** cos θ)**, put this in equation (ii)

y = u_{o}sin(θ) × x/(u_{o }cos θ) – 1/2 × g × (x/(u_{o }cos θ))^{2}OR

y = x tan θ – g x^{2}/(2u_{o}^{2}_{ }cos^{2}θ)

This is the equation of a parabolic trajectory, which describes the path followed by the projectile.

### Parabolic Motion of Projectile

This is the equation of projectile motion it is similar to the parabola **(y = ax + bx**^{2}** ) as a = tan θ **and

**b = g/(2u**

_{o}

^{2}

_{ }

**cos**

^{2}**. So, we can say that projectile motion is always parabolic in nature.**

**θ)****Applications of Projectile Motion**

**Applications of Projectile Motion**

There are various applications of Projectile Motion, some of which are as follows:

In many sports such as football, cricket, archery, shooting, rugby, etc. athletes need to have a good understanding of projectiles and their motion as this can help them predict the movement in the game and help them have an upper edge on their opponent team.Sports:

In many Military Technologies advanced projectiles such as missiles, tanks, aerial shell bombing, etc. and to make this technological advancement in the military understanding of advanced projectile motion is much needed.Military:

In many scenarios, the photographer needs to take a photo of a moving object such as sports or any other creative curiosity of the photographer. To click that perfect picture there, the photographer needs to have some understanding of the projectile motion so that he can predict the position of the object which needs to be photographed.Photography:

**Read More**

## Sample Questions on Projectile Motion

**Question 1: What is a projectile? Prove that the path of a****projectile is parabolic.**

**Solution:**

: A projectile is any object thrown into space with only gravity acting on it is called a projectile.ProjectileWe know that the equation of projectile is,

y = x tan θ – gx^{2}/2u^{2}cos^{2}comparing the equation withθy = ax + bx^{2 }Here,

a = tanθb = – g/2u^{2}cos^{2}θThe above equation of trajectory is similar to the equation of a parabola.

Hence, the path of a projectile is parabolic.

**Question 2: At what angle projectile should be projected so that the height and Range of the projectile will be equal?**

**Solution:**

If height and horizontal range will be equal,

H = R.

⇒ u

^{2}sin^{2}θ/2g = u^{2}sin2θ/g⇒ sin

^{2}θ = sin2θ [Here, sin2θ = 2 sinθ cosθ and tanθ = sinθ/cosθ]⇒ tan θ = 4

So,

θ = tan^{-1}(4)

**Question 3: Define horizontal range and find the range of a projectile thrown at 98 m/s with an angle of 30 degrees from horizontal. (use g = 9.8 m/s**^{2}**)**

**Solution: **

The horizontal distance travelled by the body performing projectile motion is referred to as the range of the projectile.Horizontal Range:Horizontal Range, R = u

^{2}sin2θ/g⇒ R = (98)

^{2}× (60°) / 9.8

⇒ R = 490√3 m

**Question 4: Name the physical quantities which will remain unchanged during the projectile motion.**

**Solution: **

The physical quantities that remain unchanged during the projectile motion are,

- Horizontal component of Velocity
- Horizontal component of Momentum
- Acceleration
- Total Energy

**Question 5: What is the maximum height attained by a ball of mass 100 g projected at an angle of 30° from the ground with an initial velocity of 11 m/s and an acceleration due to gravity of g = 10 m/s**^{2}**?**

**Solution:**

We know that the formula for maximum height is,

H = (usinθ)

^{2}/2gGiven: u = 11 m/s, θ = 30°, g = 10 m/s

^{2}Hence, Putting the values we get,

H = 1.5125 m

**Question 6: A Football is launched at a 45° angle from the ground with an initial velocity of 10 m/s; the gravity acceleration is g = 10 m/s**^{2}**. What is the time of the flight?**

**Solution:**

We know that the formula for calculating the time of flight is,

t = 2(usinθ/g)

Given: θ = 45°, u = 10m/s, g = 10m/s

^{2}Putting the values we get,

t = 1.4142 s

**Question 7: A projectile is projected from point O at an angle of 30° with an initial velocity of 30 m/s. The projectile hits the ground at point M. (Consider acceleration of gravity g = 10m/s**^{2}**) Find the following:**

**What is the total time of the flight?****What is the Horizontal Range of the projectile (OM)?****What is the maximum height of the projectile?**

**Solution:**

Given:

- Initial velocity u = 30m/s.
- Angle of projection, θ = 30°.

1. Time of FlightWe know that the total time of flight by the projectile is given by-

t = 2usinθ/g

Putting the given values,

t = 2 × 30 sin30°/10

⇒ t = 3 s

2. Horizontal RangeWe know the formula for the horizontal range is:

R = u

^{2}sin2θ/g.Putting the values we get,

R = (30)

^{2}sin60° /10⇒ R = 45 √3 m.

3. Maximum HeightMaximum height of the projectile is given by the formula:

H

_{max}= u^{2}sin^{2}θ/2gPutting the values we get,

H

_{max}= (30)^{2}sin^{2}30°/2 × 10⇒ H

_{max}= 11.25 m.

## FAQs on Projectile Motion

### Q1: What is Projectile Motion?

**Answer:**

The motion of an object under the influence of gravity when thrown in the air at some angle with the horizontal is known as projectile motion, and the object is known as projectile.

### Q2: What are Examples of Projectile Motion?

**Answer:**

Some examples of projectile motion include a cannonball being fired from a cannon, a rock thrown into the air, ball hit by batsmen in cricket, etc.

### Q3: What are the Key Factors that Affect Projectile Motion?

**Answer:**

The key factors that affect the motion of the projectile are initial velocity, angle with horizontal at which projectile is launched, and the gravitation force acting on the object.

### Q4: What is the Difference between Horizontal and Vertical Motion in Projectile Motion?

**Answer:**

The key difference between horizontal and vertical motion that in projectile motion, horizontal motions remains constant and vertical motion remains changing under the influence of gravity.

### Q5: How to Calculate the Maximum Height and Range of a Projectile?

**Answer:**

Maximum Height of the projectile is given by (H

_{max}):

Maximum Height (H_{max}) = u^{2}sin^{2}θ/2gHorizontal Range of the projectile is given by (R):

Horizontal Range (R) = u^{2}sin2θ/g

### Q6: Does the Initial Velocity of a Projectile Affect its Trajectory?

**Answer:**

Yes, Initial velocity effect the trajectory of the projectile as equation of trajectory is given as follows:

y = x tan θ – g x^{2}/(2u_{o}^{2}_{ }cos^{2}θ)Where

uis the initial velocity of the projectile._{o }

### Q7: Does the Angle of the Launch of a Projectile Affect its Trajectory?

**Answer:**

Yes, angle of the launch of a projectile affects its trajectory as equation of trajectory is given by

y = x tan θ – g x^{2}/(2u_{o}^{2}_{ }cos^{2}θ)Where

is the angle of projectile from the horizontal.θ