Thrown Object

1. Aug 4, 2004

kishtik

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

2. Aug 4, 2004

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.

3. Aug 4, 2004

kishtik

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

4. Aug 4, 2004

arildno

5. Aug 4, 2004

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

6. Aug 5, 2004

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

7. Aug 5, 2004

jtolliver

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 thats 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.