The equation of the **trajectory of a projectile** shows that this trajectory path represents a parabola. This means that a projectile follows a parabolic path.

**Equation of Trajectory of a Projectile**

Referring to Figure 1 above, we can calculate and get the

trajectory path of a projectileas follows.Equation of Trajectory of a Projectile:y = (tanθ) x – (1/2) g. x^{2}/(V_{0}cosθ)^{2}

As you see this can be rewritten in the form **y = ax + bx ^{2}** where a and b are constants.

This is again an equation that represents Parabola.

Hence we can say that trajectory of a projectile will follow a curve that is represented as

Parabola.

Deriving the **Equation of the Trajectory of a Projectile**

Referring to Figure 1 above, the initial velocity component along X-axis = V_{0x} = V_{0} cosθ and the initial 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 the X and Y axes.

**At time T=t,** (i.e., for any time instant t)

Displacement along X-axis = x= V_{0x}.t = (V_{0} cosθ). t ………… (1) and

Displacement along Y-axis = y = (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}or, y = (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 + bxwhere a and b are constants. This is an equation representing a parabola.^{2}

So we can say that the trajectory or motion path of a projectile is a parabola.

So once thrown at an angle (excluding theright angle) with the horizontal, a projectile will follow a curved path named Parabola.