Question

How does projectile motion follow a parabolic path?

Answer 1

Suppose a projectile of mass ‘m’ is projected with velocity ‘u’ at an angle θ with the ground.

The horizontal component of the initial velocity is, u_x = u cosθ [this remains constant]

The horizontal displacement at any time ‘t’ is, X = u_x t = ut cosθ

=> t = X/(u cosθ)

The vertical component is given by, u_y = u sinθ

The vertical displacement at any time ‘t’ is, Y = u_y t – ½ gt ² = ut sinθ – ½ gt²

=> Y = u[X/(u cosθ)]sinθ – ½ g[X/(u cosθ)]²

=> Y = X tanθ – gX² /[2u² cos² θ]

Which is the equation of a parabola. So, the path of the projectile is parabolic.