# Trajectory of projectile with friction in air

1. Jun 4, 2007

### chaoseverlasting

1. The problem statement, all variables and given/known data
This is not a hw prob, its just something I was trying out.

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

2. Relevant equations

F=-bv, where v is a vector.

3. 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?

2. Jun 4, 2007

### 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.

Last edited: Jun 4, 2007
3. Jun 4, 2007

### andrevdh

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

4. Jun 4, 2007

### ice109

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

$$a_x(t)=k(\frac{dx}{dt})^2$$
$$\frac{d^2(x)}{dt^2}=k(\frac{dx}{dt})^2$$

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

Last edited: Jun 4, 2007
5. Jun 5, 2007

### 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?

6. Jun 5, 2007

### 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

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

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

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.

Last edited: Jun 5, 2007
7. Jun 5, 2007

### chaoseverlasting

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

8. Jun 6, 2007

### 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.