Does air friction make a projectile trajectory non-parabolic?

[kishtik]

How can I find the trajectory of a projectile when not neglecting air friction? Will it still be a parabola?

 

[arildno]

1. There is no closed form exact solution to this problem if you assume that the air resistance has a quadratic dependence on the velocity.
(You'll need to use either numerical methods or smart approximations)
2. You may find a closed form solution in the case of a linear dependence in velocity of the air resistance.
3. No, the curves will not be parabolas.

 

[kishtik]

But how can I use the quadratic formula to find the trajectory?

 

[arildno]

What quadratic formula?

 

[Nenad]

no you can't use quadratic formula, since the path taken by the projectile will not be parabolic.

 

[arildno]

Are you talking of solving for the trajectory with qudratic dependency of velocity in air resistance?
If so, use for example a forward Euler scheme with a standard iteration loop to handle the nonlinearity

 

[jtolliver]

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.