Projectile Motion with Air Friction: Parabola or Not?

AI Thread Summary
When considering projectile motion with air friction, the trajectory will not be a parabola due to the nonlinear effects of air resistance. There is no exact closed-form solution for cases where air resistance is quadratically dependent on velocity; numerical methods or approximations must be used instead. A closed-form solution is possible with linear air resistance, but the resulting equations produce exponential terms rather than quadratic ones. The equations of motion under air resistance lead to complex relationships between position and time, making it challenging to express y in terms of x. Ultimately, the trajectory deviates significantly from parabolic behavior due to the influence of air friction.
kishtik
Messages
96
Reaction score
0
How can I find the trajectory of a projectile when not neglecting air friction? Will it still be a parabola?
 
Physics news on Phys.org
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.
 
But how can I use the quadratic formula to find the trajectory?
 
What quadratic formula?
 
no you can't use quadratic formula, since the path taken by the projectile will not be parabolic.
 
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
 
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.
 
Back
Top