Question

Derive an equation for the path of a projectile fired parallel to horizontal

Answer 1

in horizontal projected particle ,
acceleration in horizontal =0
acceleration in vertical = -g
let initial velocity in horizontal =U

★ velocity doesn't change in horizontal
★ velocity of vertical may change
now,
kinematics equation for horizontal
X = Uxt + 1/axt²
X = Uxt +0
X = Ut --------(1)
kinematics equation for vertical
Y = Uyt + 1/2ayt²
Y = 0 -1/2gt²
put equation here

Y = -1/2g(X/U)²

Y = -gX²/2U²

Answer 2

Acceleration in vertical direction = a_y = - g
      Initial velocity (vertical) = u_y = 0
So     v_y  = u_y - g t = - g t                  (∵ kinematics   v = u + a t) 
and     y = u_y t - 1/2 g t²  = -1/2 gt²                    (∵ s = ut +1/2 a t²)
              
Acceleration in horizontal direction = a_x = 0.  
      So horizontal velocity = v_x = constant = u (speed of projection)
So       x = u t 
    =>  t = x/u

=>    y = -1/2 g (x/u)² = -  x² / (2u²)

This is the path of the projectile.