Question

Derive an equation for the path of a projectile llel to horizontal

Answer 1

Ф = angle of projection wrt horizontal
initial velocity = u
x = u cos Ф  t            as there is no acceleration in x direction
         t = x/(u cos Ф)

y = u sin Ф t - 1/2 g t²
    = u sin Ф (x/ucosФ)  - 1/2 g x²/(u² cos² Ф)
y = x  tan Ф - 1/2 * g/u² x² sec² Ф
    = - (g sec²Ф /2u²) [x² - 2(u² sinФ cosФ/g)  x ]
    = - (g sec²Ф /2u²) [x - u² sinФ cosФ/g ]² + g sec²Ф/2u² *u⁴ sin²Ф cos²Ф/g²
    
(y - u² sin²Ф /2g)  = - g (sec²Ф /2u²) [x - u² sin2Ф/2g ]²

(inverted) Parabola with Vertex at  [u² sin2Ф /2g , u² sin²Ф/2g ]
Focus at :  F [ u² sin2Ф/2g,  u² sin²Ф/2g - g sec²Ф/8u² ]

Answer 2

let any projected with velocity u and inclination angle with horizontal is @ .
then ,
distance covered in x-axis is ---
x =ut + 1/gt^2
according to concept of projectile ,
initial velocity =ucos@
acceleration in x -axis =0
now,
[email protected]

t =x/ucos@ ----------------(1)

again ,
distance covered in y-axis is ------
y =ut + 1/at^2
in y -axis initial velocity =usin@
acceleration = -g
so,
y = [email protected] - 1/2 gt^2 ----------------------(2)

put equation (1) in equation (2)
y [email protected]/ucos@ -1/2g {x/ucos@}^2

y [email protected] - gx^2/2u^2cos^2@