**Projectile Motion** **Derivation**: When an object is in flight after being projected or thrown then that object is called a **projectile **and this motion under the influence of constant velocity along the horizontal and downward gravitational pull along the vertical is called **Projectile Motion**.

We will discuss in detail as well as focus on Derivation of **Projectile Motion Equations** and find out how the motion path looks like under the influence of both horizontal and vertical components of the projectile velocity.

When we throw some object making an angle with the horizontal (angle less than 90 degrees), then what is/are the force/s acting on that object?

As I throw a stone like this and it goes off my hand then I am no more applying any force on that object.

But there will be a force omnipresent acting on that stone. That is Gravity or Gravitational force applied by the earth. And this force directs vertically downwards pointing to the center of the earth.

If gravity was not present that stone would continue to move in a straight line maintaining the same angle with the horizontal plane. But this downward pull of gravity takes it down.

On one hand, that stone tries to maintain a constant velocity along the horizontal as there is no force on it along that direction. On the other hand, it gets a downward pull caused by gravity and faces an acceleration equal to g (acceleration due to gravity). Under these 2 movements,** it takes a specific path, which we will discuss now.**

## Projectile & Projectile Motion – definitions

**Projectile definition**: When an object is in flight after being projected or thrown then that object is called a **projectile **and this motion is called** Projectile Motion**. Example, the motions of a cricket ball, baseball.

**Projectile motion definition**: The motion of a projectile is known as **projectile motion** which is a result of 2 separate simultaneously occurring components of motions. One component is along a horizontal direction without any acceleration (as no force acting in this direction) and the other along the vertical direction with constant acceleration due to the force of gravity. (considering air resistance as nil)

**Projectile Motion Equations**: **In the next sections** we will discuss and derive a couple of **projectile motion equations**.

**Projectile Motion** **Derivation** –Projectile Motion Equations

We will cover here **Projectile Motion** **Derivation** to derive a couple of equations like the

* **the projectile path equation**

* **equation for time to reach maximum height**

* *total time*

* Maximum* height* of a projectile and

* **equation for th e horizontal range of a projectile**

## Projectile Motion equations derivation – parabola

**Projectile Motion** **Derivation** – Say an object is thrown with uniform velocity V_{0} making an angle theta with the horizontal (X) axis.

The velocity component along X-axis = V0x = V0 cosθ and the velocity component along Y-axis = V_{0y} = V_{0}sinθ. **Air resistance is taken as negligible.**

At time T = 0 , there is no displacement along X and Y axes. So X_{0}=0 and Y_{0}=0

**At time T=t,****Displacement along X-axis = x= V _{0x}.t = (V_{0} cosθ). t …………….. (1) and**

**Displacement along Y-axis = y= V**

_{0y}.t = (V_{0}sinθ ).t – (1/2) g t^{2}…………(2)From equation 1 we get: t = x/(V_{0} cosθ) ……………….. (3)

Replacing t in equation 2 with the expression of t from equation 3:

**y = (V _{0}sinθ ). x/(V_{0} cosθ) – (1/2) g [ x/(V_{0} cosθ)]^{2} **

**= (tanθ) x – (1/2) g . x**

^{2}/(V_{0}cosθ)^{2}………………….. (4)In the above equation g, θ and V_{0} are constant.

So rewriting equation 4:

y = ax + bx^{2} where a and b are constants.

This is an equation representing parabola.

So we can say that the **motion path of a projectile is a parabola. So once thrown a projectile will follow a curved path named Parabola.**

## Time to reach the Maximum Height by a projectile

When the projectile reaches the maximum height then the velocity component along Y-axis i.e. Vy becomes 0. Say the time required to reach this maximum height is t_{max} . The initial velocity for the motion along Y-axis (as said above) is V_{0}sinθ

Considering vertical motion along y-axis:

V_{y} = V_{0}sinθ – g t_{max}

=> 0 = V_{0}sinθ – g t_{max}

=> **t _{max}= (V_{0}sinθ )/g ……………………………..(5)**

So this is the equation for the time required to reach the maximum height by the projectile.

**Total time of flight for a projectile:**

**So to reach the maximum height by the projectile the time taken is (V _{0}sinθ )/g**It can be proved that the projectile takes equal time [ (V

_{0}sinθ )/g] to come back to the ground from its maximum height.

Therefore the **total time of flight for a projectile T _{max}** =

**2(V**

_{0}sinθ )/g …………………. (6)## Maximum Height reached by a projectile

Let’s say, the maximum height reached is H_{max} . We know that when the projectile reaches the maximum height then the velocity component along Y-axis i.e. V_{y} becomes 0.

Using one of the motion equations, we can write

(V_{y})^{2} =( V_{0}sinθ )^{2} – 2 g H_{max}

=> 0 = ( V_{0}sinθ )^{2} – 2 g H_{max}

=> **H _{max} = ( V_{0}sinθ )^{2}/(2 g) …………………… (7)**

## Horizontal range of a projectile

Say R is the horizontal range covered by a projectile is R when it touches the ground after passing the time **T _{max}**

Here we will use the horizontal component of the velocity only as the effective velocity to traverse this horizontal path.

R = (V_{0} cosθ) . **T _{max}** = (V

_{0}cosθ).

**2(V**= (

_{0}sinθ )/g**V**

_{0}^{2 }sin2θ )/ g**R = (V _{0}^{2 }sin2θ )/ g ………………………. (8)**

For the Maximum value of R, it is pretty easy to understand from the above equation that the angle θ needs to be equal 45 degrees, because in that case, we get the maximum value from sine as 2θ becomes 90 degrees**.**

Hence **R _{max} = V_{0}^{2 }/ g …………………………. (9)**

So we have discussed the Projectile Motion Derivation steps to derive a set of projectile motion equations. We will add some numerical as well very soon.

Related study: **Equations for free fall for vertical motion**