Trajectory of projectile with friction in air

[chaoseverlasting]

Homework Statement

This is not a homework prob, its just something I was trying out.

To find the equation of the path of the projectile when air friction is considered.

Homework Equations

F=-bv, where v is a vector.

The Attempt at a Solution

Let the projectile be launched with a velocity v=v_{0x}i+ v_{0y}j+v_{0z}k;

Velocity after a time t is: v(t)=\frac{mv_{0x}}{m+bt}i + \frac{(v_{0y}-gt)m}{m+bt}j + \frac{mv_{0z}}{m+bt}k. Integration of this expression wrt t, would give you the path of the projectile... right?

 

[ice109]

i don't know how you got your function but i know mass usually cancels out

if you treat friction as a constant force, independent of velocity and time, though I'm pretty friction force in a fluid is not independent of velocity, you just get the regular position function in each component with an acceleration because

a_x,z(t)=-k and in the x direction you just get a bigger acceleration a_y(t)=-k+-g k and g being both constants you would just get another constant acceleration, bigger.

 

[andrevdh]

I think at higher speeds it is proportional to the square of the speed.

 

[ice109]

I think at higher speeds it is proportional to the square of the speed.

im bad at this so i would like to test my grasp as well

<br /> <br /> a_x(t)=k(\frac{dx}{dt})^2
<br /> \frac{d^2(x)}{dt^2}=k(\frac{dx}{dt})^2<br /> <br />

?? would this be the correct equation for acceleration for the x component?
i will try solving the DE if this is correct

 

[chaoseverlasting]

Err... ok, I took friction to be -bv (vector), so the acceleration in the x and z directions worked out to be \frac{-b}{m}v_z. Since the velocity would change with time as:

v_z=v_{0z}-\frac{b}{m}v_zt, this would give v_z to be:

\frac{mv_{0z}}{m+bt}, where the mass does cancel out dimensionally. Integration of this expression gives a logarithmic function... so, is this right?

 

[andrevdh]

The air resistance, R, opposes the velocity of the projectile.

The angle of attack, \theta, of the projectile changes in its trajectory.

By resolving in the x - and y directions we get that

m\ddot{x} = - R \cos(\theta)

and

m\ddot{y} = - R \sin(\theta) - mg

we can rewrite the sine and cosine terms as

\cos(\theta) = \frac{\dot{x}}{v}

and

\sin(\theta) = \frac{\dot{y}}{v}

leading to

\ddot{x} = -a\dot{x}v}

and

\ddot{y} = -a\dot{y}v} - g

An "easier" approach is to rather resolve along the tangential and normal directions to the trajectory - which leads to the equation of the hodograph.

 

[chaoseverlasting]

Yeah, that's exactly what I've done... just used pure vectors in 3D instead of resolving the force...

 

[andrevdh]

I've burned myself once before with this 3D vector stuff, so I would rather keep to what I know - resolving the force components and setting up the equations of motion.