Question

show the path of projectile in parabola

Answer 1

projected with velocity ‘u’ at an angle θ with the ground.
The horizontal component of the initial velocity is, ux = u cosθ [this remains constant]
The horizontal displacement at any time ‘t’ is, X = ux t = ut cosθ
=> t = X/(u cosθ)
The vertical component is given by, uy = u sinθ
The vertical displacement at any time ‘t’ is, Y = uy 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.

Answer 2

See the above attachment to see the solution.