When an object is launched, it follows a parabolic path and the motion known as projectile motion. This post will look at the parameters and how to calculate projectile motion in a detailed analysis.

**When an object is launched and moves along a symmetrical parabolic path, the motion is referred to as projectile motion.The object’s parabolic path is referred to as its trajectory. The object, in this case, travels vertically and horizontally at the same time. As a result, projectile motion is two-dimensional. In projectile motion, you only need to provide force at the start of the trajectory; the object is only affected by gravity after that. **

**Now let us see how to calculate projectile motion:**

Assume you’re shooting a cannonball. It begins to go upward and forward until it reaches its maximum height. From here on out, it will continue to go forward but in a downward direction.** It is tracing this curved route known as a trajectory, which has the form of a parabola. Any object moving in this manner is referred to as being in projectile motion. As the path of projectile motion is always parabolic, it is represented as:**

**y = ax + bx ^{2}**

Before reaching the earth, the cannonball will take a parabolic route during its journey. **The velocity along the X-axis remains constant throughout the motion, whereas the velocity along the Y-axis varies with its position. Only the acceleration due to gravity, 9.8 m/ s ^{2}, governs this type of motion. The acceleration pointing down remains constant during the flight of cannonballs. **

**Kinematic Equations for Projectile Motion:**

**Initial Velocity formula:**

Please assume that the initial velocity is u and the projectile angle is 𝛳. **There are two components to the initial velocity: horizontal and vertical.**

The horizontal component of initial velocity is u_{x} and given by:

u_{x} = u ᐧ cos𝛳

And the vertical component of initial velocity is uy and given by:

u_{y} = u ᐧ sin𝛳

**Time of Flight of projectile:**

**The time of flight in projectile motion is the time gap between the object being launched and reaching the ground. The magnitude of the starting velocity and the angle of the projectile define the time of flight, which is denoted by T.**

**Acceleration formula:**

**There is no acceleration in the horizontal direction because the horizontal component of acceleration remains constant throughout the motion. The only acceleration in the vertical direction is due to gravity.**

a_{x} = 0 and

a_{y} = -g

The negative sign stands for downward acceleration.

**Velocity at time ‘t’ formula:**

**Throughout the motion, the horizontal component of velocity will remain constant. However, because vertical acceleration is constant, the vertical component of velocity varies linearly.**

As a result, velocity can be calculated at any time t using the following formula:

v_{x} = u_{x} = u ᐧ cos𝛳

v_{y} = u ᐧ sin𝛳 – g ᐧ t

Using Pythagoras Theorem, one can find the magnitude of velocity.

**Displacement at time ‘t’ formula:**

At time t, displacement can be given by:

x = (u ᐧ cos𝛳) ᐧ t

y = (u ᐧ sin𝛳) ᐧ t – ½(gt^{2})

**Parabolic Trajectory formula:**

We can utilize the displacement equations in the x and y directions to derive an equation for the parabolic shape of a projectile motion:

**Range of Projectile formula:**

**The total horizontal distance traveled by the object during its flight time is defined as its range.** If the object is being launched from the ground (starting height = 0), the formula is as follows:

According to the equation above, the maximum horizontal range can be obtained when the projectile angle 𝛳 = 45°. Rm represents the maximum range.

**The Maximum Height formula:**

**When the vertical velocity component is zero, v _{y} = 0, the maximum height can be attained. Because the time of flight is the total time for the projectile, it will take half of that time to achieve maximum height.** Thus, time to reach a maximum height is,

Thus, from the displacement equation, the maximum height can be given by:

**Horizontal Projectile Motion Formula:**

**Horizontal projectile** **motion is a type of projectile motion in which an object is launched horizontally from an elevated plane rather than from the ground. **

The launch angle does not need to be specified because it is parallel to the ground (i.e., the angle is 0°). As a result, we only have one initial velocity component: Vx = V, whereas Vy = 0.

## In this case, the equations of motion are as follows:

**Velocity of Horizontal Projectile motion:**

Horizontal velocity: v_{x} = v

And vertical velocity: v_{x} = -g ᐧ t

**Distance traveled by object in Horizontal projectile motion:**

In this case, the horizontal distance is calculated as follows:

x = v ᐧ t

And the vertical distance can be given by:

y = -(g ᐧ t^{2}) / 2

**Acceleration in Horizontal projectile motion:**

Horizontal acceleration ax = 0, as horizontal velocity is constant.

Vertical acceleration a_{y} = -g

**Trajectory Equation of Horizontal projectile motion:**

Trajectory equation, in this case, can be given by:

**Time of flight in Horizontal projectile motion:**

Time of flight, in this case, can be given by:

**Range of the projectile in Horizontal projectile motion:**

The range of projectile in horizontal projectile motion is:

Because we are launching the object from maximum height, we do not need to calculate maximum height in this scenario.

Let us look at some projectile motion problems.

**Problem 1: What will be the θmax for which the distance of the particle from the thrower always increases up to the end of the path again at the ground?**

**Solution:** The horizontal distance traveled by an object is called its horizontal range and is given by:

Maximum range can be achieved when the projectile angle is 45°.

Thus, for Rm maximum angle θmax = 45°.

**Problem 2: If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is :**

**Solution:** As the ball is thrown vertically, projectile angle 𝛳 = 90°.

As 𝛳 = 90°

Where Tm is the time taken by an object to reach maximum height.

Assume that h represents the distance travelled by an object during the final t seconds of its ascent. The velocity at that moment is then calculated as follows:

V = u – g ᐧ (T – t)

= u – g ᐧ (u/g – t)

= gt

Thus distance covered in last t second is :

h = vt – ½ gt^{2}

= gt^{2} – ½ gt^{2}

= ½ gt^{2}

**Problem 3:** **A particle is projected at an angle of 60° above the horizon with a speed of 10 m/s. After some time, the velocity makes an angle of 30° above the horizontal. The speed of the particle at this instant is?**

**Solution:** The horizontal component of velocity is given by:

v_{x} = u ᐧ cos𝛳

Here in the first case the angle of projection is 60° and initial velocity u = 10 m/s. Thus,

v_{x} = u ᐧ cos60

= 10 x 0.5

= 5 m/s.

Now, the vertical component of velocity v_{y} changes during motion, but v_{x} remains constant. Thus,

v_{x} = v ᐧ cos𝛳2

Where 𝛳2 = 30° and v is velocity when an object makes angle 𝛳 = 30° with the horizon.