The motion of object in two dimensions is explained by two main principles that are kinematic principles and Newton’s laws of motion. The motion in two dimensions is called the motion of projectiles. The projectile object is one on which is moved under the force of gravity. Some examples of projectiles are like an object is dropped from its rest condition on which there is no effect of air resistance. An object is thrown from vertical under negligible effect of air resistance an object and if it is thrown in up direction at a horizontal angle is considered in a **projectile motion**.

So, we can say that a projectile is any object which is projected continues in motion under its inertia and force of gravity.

**Projectile Motion Definition:**

**Projectile motion is an example of curved motion with constant acceleration. It is two dimensional motion of a particle thrown obliquely into the air.**

Consider the motion and path followed by the ball when it moves in the curved path. We will make** **two assumptions** **here:

a) First assumption is that the free fall acceleration (g) remains constant and does not change its value during the motion of the ball.

b) Resistance offered by the ball is negligible.

**Projectile Motion Formula:**

We often experience many kinds of motions in our daily life. **Projectile motion** is one among them. A projectile is some object thrown in air or space. The Curved path along which the projectile travels is what is known as trajectory.

**Projectile Motion** is the free fall motion of any body in a horizontal path with constant velocity.

** **Projectile Motion Formula **(trajectory formula) **is given by:-

Horizontal distance, x = v_{x}t

Horizontal Velocity, v_{x} = v_{xo}

Vertical Distance, y = v_{yo} t – ½ gt²

Vertical Velocity, v_{y} = v_{yo} – gt

Where,

v_{x} = Velocity along x – axis,

v_{xo} = Initial velocity along x – axis,

v_{y} = Velocity along y – axis,

v_{yo} = Initial velocity along y – axis.

Equations related to trajectory motion (projectile motion) are given by,

Time of flight, t = \(\frac{2{{v}_{0}}\,\sin \,\theta }{g}\)

Maximum height reached, H = \(\frac{{{v}_{0}}^{2}\,{{\sin }^{2}}\theta }{2g}\)

Horizontal range, R = \(\frac{{{v}_{0}}^{2}\,\sin \,2\theta }{g}\)

Where,

v₀ = Initial velocity,

Projectile Motion formula is used to find the distance, velocity and time taken in the projectile motion.

**How to find the Projectile Motion?**

**Example: **A body is projected with a velocity of 20 m/sec at 50⁰ to the horizontal. Find the

- Maximum height reached
- Time of flight and
- Range of the projectile.

**Solution: **Initial Velocity, v₀ = 20 m/sec

θ = 50⁰

Time of flight, t = \(\frac{2{{v}_{0}}\,\sin \,\theta }{g}\)

= \(\frac{2\times 20\times \sin \,{{50}^{0}}}{9.8}\)

= 3.126 sec

Maximum height reached, H = \(\frac{{{v}_{o}}^{2}\,{{\sin }^{2}}\theta }{2g}\)

= \(\frac{{{(20)}^{2}}\,{{\sin }^{2}}\,{{50}^{2}}}{2\times 9.8}\)

= 11.97 m

Horizontal Range, R = \(\frac{{{v}_{0}}^{2}\,\sin \,2\theta }{g}\)

= \(\frac{{{20}^{2}}\,\sin \,{{100}^{0}}}{9.8}\)

= 40.196 m