### Projectile motion and Projectile motion equations-ready reference

##### November 30, 2017

# Projectile & Projectile Motion – definitions

In this post we would define and discuss on **Projectile** and **Projectile Motion**. We will also derive **Projectile Motion Equations **and find out how the motion path looks like under the influence of both horizontal and vertical component of the projectile velocity.

**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 cricket ball, base ball.

**Projectile motion definition**: The motion of a projectile is known as **projectile motion** which is basically 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 couple of projectile motion equations.

## Projectile Motion Equations

We will cover here 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 horizontal range of a projectile

### Projectile Motion equation derivation – parabola

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 = V_{0x} = V_{0} 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 as 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. V_{y} 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 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 Maximum value of R , it is pretty easy to understand from the above equation that the angle **θ **needs to be equal 45 degree, because in that case we get the maximum value from sine as **2****θ becomes 90 degree.**

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

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

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