

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.
(Image will be uploaded soon)
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.
(Image will be uploaded soon)
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.
(Image will be uploaded soon)
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.
FAQs on Projectile Motion
1. What is projectile motion? Provide some real-world examples.
Projectile motion is the motion of an object that is thrown or projected into the air and then moves under the sole influence of gravity. The path it follows is called a trajectory. We assume that air resistance is negligible for basic calculations. Common examples include:
- A football being kicked.
- A javelin or shot put thrown by an athlete.
- A ball thrown towards a basket.
- Water jetting out of a fountain.
2. What are the main types of projectile motion studied in Class 11 Physics?
In the context of the CBSE syllabus, projectile motion is typically categorised into two main types based on the initial projection:
- Horizontal Projectile Motion: This occurs when an object is thrown with an initial velocity purely in the horizontal direction, for example, a ball rolling off a tabletop.
- Oblique (or Angular) Projectile Motion: This is when an object is thrown at an angle to the horizontal (other than 90°). This is the most general case, like a cricket ball hit for a six. A vertical throw (90°) is a special, one-dimensional case.
3. What is the difference between a projectile and its trajectory?
The terms are related but distinct. The projectile is the actual object that is in motion (e.g., the cannonball, the stone, the baseball). The trajectory, on the other hand, is the imaginary path or curve that the projectile follows through space (which is typically a parabola under ideal conditions).
4. Why is the path of a projectile considered a parabola?
The trajectory of a projectile is a parabola due to the combination of two independent motions:
- Horizontal Motion: The object moves with a constant horizontal velocity because there is no acceleration in the horizontal direction (assuming no air resistance).
- Vertical Motion: The object experiences constant downward acceleration due to gravity (g).
5. What are the key assumptions made when analysing projectile motion?
To simplify the analysis of projectile motion as per the NCERT curriculum, several key assumptions are made:
- Air resistance is negligible: This is the most significant assumption. In reality, air drag affects the object's speed and path.
- The acceleration due to gravity (g) is constant: We assume 'g' (approx. 9.8 m/s²) is constant in both magnitude and direction throughout the motion.
- The Earth's curvature is ignored: The ground is considered a flat plane.
- The Earth's rotation is ignored: We do not account for effects like the Coriolis force.
6. How do the horizontal and vertical components of a projectile's velocity change over time?
The two components behave very differently:
- The horizontal velocity (vₓ) remains constant throughout the flight. This is because there are no horizontal forces acting on the projectile (as we ignore air resistance).
- The vertical velocity (vᵧ) continuously changes due to gravity. It decreases as the projectile moves upwards, becomes zero at the highest point of the trajectory, and then increases in the downward direction.
7. Is the total mechanical energy of a projectile conserved? Explain why.
Yes, in an ideal projectile motion scenario (where air resistance is ignored), the total mechanical energy is conserved. This is because the only force doing work on the projectile is gravity, which is a conservative force. As the projectile's height changes, its potential energy and kinetic energy convert into one another, but their sum remains constant at all points along the trajectory.
8. How does the angle of projection affect the range and maximum height of a projectile?
The angle of projection significantly impacts both the horizontal range (R) and maximum height (H):
- Maximum Height (H): The height increases as the angle of projection increases from 0° to 90°. It is greatest when the object is thrown straight up (90°).
- Horizontal Range (R): The range is zero for angles of 0° and 90°. For a given initial speed, the maximum horizontal range is achieved at an angle of 45°. Furthermore, for any two complementary angles of projection (e.g., 30° and 60°), the horizontal range is the same.

















