Most of you have read projectiles and know how to find the equation of trajectory when the initial velocity, angle of projection are given.

Lets consider a reverse problem:
Given a general quadratic equation, lets find its initial velocity, angle of projection, time of flight, etc., etc, etc ....

Consider a quadratic equation
y = ax²+bx+c -------------(1)
The X axis is considered to be the horizontal plane, and the Y axis is vetial, positive up.
Now, for this equation to represent a projectile, it should be a parabola that opens downwards, ie, a<0
Also, it should have two real roots. For the case of a single real root, the equation represents a particle thrown vertically upwards.
So = b² - 4ac >0

Lets assume that the particle is projected from the X axis only.
Then, if c>0, the particle is projected from the left of the origin.
If c = 0, the particle is projected from origin.
If c< 0, the particle is projected from the right of the origin.

Differentiating the equation wrt x,
dy/dx = 2ax + b
Equating it to zero, we get the critical point
x_c = -b/2a
d² y/dx²  = 2a <0
Hence x_c is a point of maxima.
H = y_max = y(x_c ) = -/4a -----------------(2)

Differentiating the equation wrt time t,
dy/dt = 2ax(dx/dt) + b(dx/dt) ---------------(3)
Differentiating again wrt t
d² y/dt² = (2ax + b)(d² x/dt²) + 2a(dx/dt)² ---------------(4)
Now,
d² y/dt² = -g & d² x/dt² = 0
Substituting these values in (4)
-g = (2a)v_x²
v_x =  (-g/2a), which is a constant. -----------------(5)
So the X component of velocity remains constant.

From (3),
v_y = (2ax+b)v_x -------------------(6)

Let the roots of the equation (1) be & such that <. Then x = is the projection point. Hence at the projection point,

u_x =  (-g/2a) &
u_y = (2a+b)v_x

Now as is a root of the equation (1)
= (-b+ )/2a
(2a+b) =

Hence at projection point,
u_x =  (-g/2a) &
u_y = (-g/2a)

So initial speed,
u = u_x² + u_y²
    =  (1+)(-g/2a) --------------------(7)

Angle of projection
= tan^-1 (u_y / u_x) = tan^-1 ------------------(8)

Time of flight
T = 2u_y/g = (-/2ag)

Horizontal range
R = 2u_y u_x /g = - /a ------------------(9)

These derived equations provide a complete description of the projectile.
u =  [(1+)(-g/2a)]
= tan^-1
R = - /a
T =  (-/2ag)
H = -/4a

Lets take an example to see these formulas in work

Let the equation be
y = -x²+5x-6

The constant term is -ve, hence the particle is projected from the  +ve X axis
=  1

u =  [(1+)(-g/2a)] = 10 m/s
= tan^-1 = 45 deg
R = - /a = 1 m
T =  (-/2ag) = 1/(2 5) s
H = -/4a = 1/4 m