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# Projectile Motion

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## What is Projectile Motion?

Two-dimensional motion is the motion of a body along a plane. If the motion of the plane is considered along the XY plane, then the x coordinate and y coordinate of the position of the particle changes with time. In two-dimensional motion, the body moves along a curved path instead of a straight path. The two main examples of two dimensional motion are projectile motion and circular motion. Here, we will discuss projectile motion in Physics, which is a two-dimensional motion.

### Projectile Motion of a Body Under Gravity

In Physics, projectile motion is defined as the motion of an object under the influence of gravity. A body thrown vertically upwards is an example of projectile motion. For projectile motion, the air resistance acting on the body is neglected and the only force acting on the body is the gravitational force. A body thrown with an initial velocity at an angle with the horizontal is a projectile motion.

In this projectile, the body moves along the vertical plane, and the path followed by the body is called a trajectory. A ball that is thrown horizontally from the top of a cliff is also a projectile motion. These are examples of projectile motion.

For a projectile motion, the gravitational force acting on the body is in the downward direction. Therefore, the body in a projectile motion has acceleration due to gravity 9.8 m/s2. It must be noted that no force is acting on the body in the projectile motion in the horizontal direction. Therefore, the acceleration of the body in the horizontal direction is zero. The projectile motion is uniformly accelerated motion. Therefore, we can apply the equations of motion in the vertical and horizontal directions. There are three equations of motion for uniformly accelerated motion as follows:

$\Rightarrow v = u+ at$

$\Rightarrow S = ut + \frac{1}{2} at^{2}$

$\Rightarrow v^{2} = u^{2} + 2as$

Where,

S- Displacement of the body

u- Initial velocity of the body

v- Final velocity of the body

t- Time taken for the motion

a- Acceleration of the body

## Types of Projectile Motion

There are different types of projectile motion based on the direction of the initial velocity of the projectile. The three main types are vertical projectile motion, horizontal projectile motion and oblique projectile. Let us learn them in detail.

### Vertical Projectile Motion

In the vertical projectile motion, the angle between the initial velocity and the horizontal direction is 90°. The body moves in a vertical direction under the action of gravity.

The horizontal initial velocity (ux) of the projectile is zero and the initial velocity is only along the vertical direction.

For a vertical projectile motion,

$\Rightarrow u_{x} = 0$

$\Rightarrow u_{y} = u$

$\Rightarrow a_{x} = 0$

$\Rightarrow a_{y} = -g$

Where,

$u$ - Initial velocity of the body

$u_{x}$ - Horizontal component  initial velocity

$u_{y}$ -  Vertical component  initial velocity

$a_{x}$ - Horizontal component acceleration of the body

$a_{y}$ - Vertical component  of acceleration of the body

The maximum height of the vertical projectile is given by the formula,

$\Rightarrow H = \frac{u^{2}}{2g}$

Where,

H- The maximum height of the projectile motion

u- Initial velocity of the body

g- Acceleration due to gravity

### Horizontal Projectile Motion

In horizontal projectile motion, the object is thrown in the horizontal direction and the angle between the initial velocity and the horizontal is zero. Therefore, the initial velocity of the projectile is along the horizontal direction and the vertical component of the initial velocity of the horizontal projectile motion is zero. The trajectory of horizontal projectile motion is a parabola. The body moves in the horizontal direction along with vertical down motion along the parabolic path.

For a horizontal projectile motion,

$\Rightarrow u_{x} = u$

$\Rightarrow u_{y} = 0$

$\Rightarrow a_{x} = 0$

$\Rightarrow a_{y} = -g$

The projectile motion derivation for the maximum vertical displacement is given as follows.

Using equation of motion,

$S_{y} = u_{y}t + \frac{1}{2} a_{y}t^{2}$

Where,

$S_{y}$- Vertical displacement of the body

$t$- time taken during the projectile motion

$u_{y}$- Vertical component of initial velocity

$a_{y}$- Vertical component  of acceleration of the body

Substitute the values for each term in the above equation to calculate the maximum vertical displacement.

$\Rightarrow S_{y} = u_{y}t + \frac{1}{2} a_{y}t^{2}$

$\Rightarrow S_{y} = 0 \times t + \frac{1}{2} \times g \times t^{2}$

$\Rightarrow S_{y} = \frac{1}{2} gt^{2}$

Therefore, the formula to calculate the maximum vertical displacement of the horizontal projectile motion is given by,

$\Rightarrow H = \frac{1}{2} gt^{2}$

### Oblique Projectile Motion

In oblique projectile motion, the initial velocity of the body is making an angle with the horizontal. The angle made by the initial velocity with the horizontal is greater than 00 and greater than 90°. Oblique projectile motion is also called angular projectile motion. The trajectory of angular projectile motion is a parabolic path. The horizontal of the initial velocity and the vertical component of the initial velocity is not equal to zero.

For angular projectile motion,

$\Rightarrow u_{x} = u sin(\theta)$

$\Rightarrow u_{x} = u cos(\theta)$

$\Rightarrow a_{x} = 0$

$\Rightarrow a_{y} = -g$

Where,

$u$ - Initial velocity of the body

$u_{x}$ - Horizontal component  initial velocity

$u_{y}$ -  Vertical component  initial velocity

$a_{x}$ - Horizontal component acceleration of the body

$a_{y}$ - Vertical component  of acceleration of the body

$\theta$ - The angle between initial velocity and the horizontal

The formula to calculate the maximum height of the projectile motion is given by,

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

The time of flight of an equation is the total time taken during projectile motion. The formula to calculate the time of flight of a projectile motion is given by,

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

The range of a projectile motion is the horizontal displacement of the body when it comes back to the horizontal surface. The formula to calculate the range of the projectile motion is given by,

$\Rightarrow = \frac{u^{2} sin2 \theta}{g}$

### Solved Examples:

1. A ball is thrown with an initial velocity of 20 m/s at an angle of 300 with the horizontal. Find the maximum height of the ball. Take acceleration due to gravity as 10 m/s2 and neglect the air resistance.

Ans:

The initial velocity of the ball = u = 20 m/s

The angle between the initial velocity and the horizontal = $\theta = 30^{0}$

The formula to calculate the maximum height of the projectile motion is given by,

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

H- Maximum height of the projectile motion.

$\theta$ - The angle between initial velocity and the horizontal

u- Initial velocity of the body

g - acceleration due to gravity

Substitute the values for each term in the above equation to obtain the maximum height of the projectile motion.

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

$\Rightarrow H = \frac{(20 m/s)^{2} sin^{2} (30^{0})}{2g}$

$\Rightarrow H = 5 m$

Therefore, the maximum height of the projectile motion is obtained as 5 m.

2. A body is given an initial velocity of 10 m/s at an angle of 150 with the horizontal. Calculate the horizontal distance covered by the body when it reaches back to the surface.

Ans:

The initial velocity of the body = u = 10 m/s

The angle between the initial velocity and the horizontal = $\theta = 15^{0}$

The horizontal distance covered by the body is the range of the projectile motion.

The formula to calculate the range of the projectile motion is given by,

$\Rightarrow R = \frac{u^{2} sin 2 \theta}{g}$

R- Range of the projectile motion.

$\theta$- The angle between initial velocity and the horizontal

u- Initial velocity of the body

g- acceleration due to gravity

Substitute the values for each term in the above equation to obtain the maximum height of the projectile motion.

$\Rightarrow = \frac{u^{2} sin2 \theta}{g}$

$\Rightarrow = \frac{(10 m/s)^{2} sin(2\times 15)}{10 (m/s)^{2}}$

$\Rightarrow = 5 m$

The range of the projectile motion is 5 m. It is the horizontal distance covered during the projectile motion

### Conclusion

Projectile motion is a two-dimensional motion. There are mainly three types of projectile motion based on the angle between initial velocity and horizontal. The projectile is a uniformly accelerated motion having acceleration equal to the acceleration due to gravity in the downward direction. The displacement and time taken during the projectile motion can be calculated by applying its equation in the horizontal and vertical directions.

Last updated date: 27th Sep 2023
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## FAQs on Projectile Motion

1. Is the total energy of a body in a projectile motion constant?

Yes, the total energy of the body in a projectile motion is a constant. In  a projectile motion, work is done by the gravitational force. Therefore, the sum of the potential energy and kinetic energy of the body in projectile motion is a constant. At different points of the projectile motion, the potential energy and the kinetic energy may vary but the total energy remains constant.

2. What are the applications of projectile motion?

There are many applications of projectile motion in the field of engineering and also in real life. In real life, the applications of projectile motions include long jumping, javelin throw etc. For an angular projectile motion, the range is maximum when the initial velocity is making an angle 450 with horizontal. So, a person participating in a high jump tries to make his initial velocity at somewhat close to 450  to get a better result. The application of projectile motion in the engineering field includes the motion of unguided rockets, bullets etc.