**Equation of motion of a horizontal projectile**

Let an object be thrown from a point O with velocity v_{0} along the horizontal direction [Figure]. If we ignore air resistance and consider g constant, then along the path of motion of the thrown object v_{0} will remain constant. Let after t sec the object reach at point P travelling distances x and y, respectively along horizontal and vertical directions. Now let the velocity at P be v. The horizontal and vertical components of v are respectively v_{x} and v_{y}.

So, **v _{x} = v_{0} = v cos θ**

and **v _{y} = v_{0} + gt = 0 + gt = v + sin θ**

[as we know, v_{0} = 0 along vertical direction]

so, **v = √(v _{x}^{2} + v_{y}^{2})**

here angle θ in between v and horizontal direction, i.e., along X-asis,

and, **tan θ = v _{y}/v_{x}**

Again, x = v_{0} + t … … … (1)

[a_{x} = 0 along horizontal direction]

so, t = x/v_{0}

and, y = ½ gt^{2} …. …. …. (2)

[v_{0} = 0 along vertical direction]

Now, inserting the value of t from equation (1) in equation (2) we get,

y = ½ g(x/v_{0})^{2}

so, **x ^{2} = (2v_{0}^{2}/g).y** … … … (3)

Since, v_{0} and g are constant, if we put 2v_{0}^{2}/g = 4A, we get,

**x ^{2} = 4A.y**

This is an equation of parabola. So, trajectory of a body or a projectile ejected horizontally in resistant free-path forms a parabola.