# Projectile motion – Derivation of Projectile motion equations for class 11

Projectile Motion Derivation: We will discuss how to derive Projectile Motion Equations or formula and find out how the motion path looks like under the influence of both horizontal and vertical components of the projectile velocity.

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.

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 projected object in flight 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)

In the next sections we will discuss and derive a couple of projectile motion equations.

## Derivation of Projectile Motion Equations

We will cover here Projectile Motion Derivation to derive a couple of equations or formula like:

1> the projectile path equation

2> equation for time to reach maximum height

3> total time

4> Maximum height of a projectile and

5> equation for the horizontal range of a projectile

Students of Class 11 from boards like ISC, CBSE and state boards will find this one useful.

## Projectile Motion equations derivation – parabola

Let’s start with Projectile Motion Derivation. Say an object is thrown with uniform velocity V0 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 = V0y = V0sinθ. Air resistance is taken as negligible.

At time T = 0 , there is no displacement along X and Y axes. So X0=0  and Y0=0

At time T=t,

Displacement along X-axis = x= V0x.t
= (V0 cosθ). t ………… (1) and

Displacement along Y-axis = y= V0y.t
= (V0sinθ ).t  – (1/2) g t2 …………(2)

From equation 1 we get: t = x/(V0 cosθ) ………….. (3)

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

y = (V0sinθ ). x/(V0 cosθ)  – (1/2) g  [ x/(V0 cosθ)]2
or, y = (tanθ) x –   (1/2) g . x2/(V0 cosθ)2………..(4)

In the above equation g, θ and V0 are constant.

So rewriting equation 4:
y = ax + bx2 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 tmax .

The initial velocity for the motion along Y-axis (as said above) is V0sinθ

Considering vertical motion along y-axis:

Vy = V0sinθ  – g tmax

=> 0 = V0sinθ  – g tmax

=> tmax= (V0sinθ )/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  (V0sinθ )/g

It can be proved that the projectile takes equal time [ (V0sinθ )/g] to come back to the ground from its maximum height.

Therefore the total time of flight for a projectile Ttot = 2(V0sinθ )/g …………………. (6)

## Maximum Height reached by a projectile

Let’s say, the maximum height reached is Hmax . We know that when the projectile reaches the maximum height then the velocity component along Y-axis i.e. Vy becomes 0.

Using one of the motion equations, we can write

(Vy)2 =( V0sinθ )2 – 2 g Hmax

=> 0 = ( V0sinθ )2 – 2 g Hmax

=> Hmax = ( V0sinθ )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 Tmax

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

R = (V0 cosθ) .  Tmax
= (V0 cosθ). 2(V0sinθ )/g = (V0sin2θ )/ g

R = (V0sin2θ )/ 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 Rmax = V0/ g  ………………….. (9)

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

## List of Projectile motion formula or equations

Motion Path equation: y = (tanθ) x –   (1/2) g . x2/(V0 cosθ)2

Time to reach max height: tmax= (V0sinθ )/g

Total time of flight for a projectile Ttot = 2(V0sinθ )/g

Maximum height reached: Hmax = ( V0sinθ )2/(2 g)

Horizontal range of a projectile: R = (V0sin2θ )/ g

Maximum possible horizontal range: Rmax = V0/ g

### State the equation of path of the projectile

Motion Path equation can be expressed as y = (tanθ) x –   (1/2) g . x2/(V0 cosθ)2 