# Prove that projectile motion is parabolic class 11

Here we will Prove that projectile motion is parabolic (class 11). To do this we need to derive the **Projectile Motion Path Equation** (or *Projectile trajectory equation*) first.

**Projectile motion is parabolic – let’s prove this (class 11)**

Say an object is thrown with uniform velocity V_{0} making an angle θ with the horizontal (X) axis.

The initial velocity component along the horizontal X-axis = V

_{0x}= V_{0}cosθ

And the initial velocity component along the vertical 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.

Let’s rewrite equation 4:

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

This is an equation representing a parabola.

So we can say that the trajectory or motion path of a projectile is a parabola. And, projectile motion is parabolic.So once thrown at an angle (excluding theright angle) with the horizontal, a projectile will follow a curved path named Parabola.