kishtik said:
But how can I use the quadratic formula to find the trajectory?
You dont.
Without air resistance the equations are:
m \frac{d^2 x}{dt^2} = 0
and
m \frac{d^2 y}{dt^2} = -mg
these can be solved by integrating both sides twice(the intitial height and initial velocity are functions of the constants of integration). The solution for x is linear in t and the solution for y is quadratic for t.
with air resistance(proportional to the velocity) the equations are:
m \frac{d^2 x}{dt^2} = -k \frac{dx}{dt}
and
m \frac{d^2 x}{dt^2} = -k \frac{dx}{dt} - mg.
If you solve these you will find exponentials turning up in the solutions, and neither x nor y is quadratic in t. The solution is
x=A+Be^{-kt/m}
y=C+De^{-kt/m}-\frac{mg}{k}t
The initial position is (A+B,C+D). The xcomponent of the initial velocity is -kB/m. The y component of the initial velocity is -kD/m - mg/k. It is somewhat difficult to find y in terms of x, but it can be done by using logarithms as follows:
x-A = Be^{-kt/m}
\frac{x-A}{B} = e^{-kt/m}
ln(x-A)-ln(B) = -kt/m
-\frac{k}{m}(ln (x-A) - ln (B)) = t
now all that's left is to substitute this in the solution for y, to obtain
y=C+D\frac{x-A}{B} +g(ln (x-A) - ln (B))
This clearly is not a parabola.