# Projectile motion + Air resistance equation?

1. Jul 1, 2005

### Thaer

Take a look at this little web-progarm
http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html

I know that it uses the following equation to calculate the range of the projectile without air resistance
R = (v^2 sin2(theta))/g
R - Range
v - intial velocity

But I have no clue how it calculates range of the projectile with air resistance.
Does anyone know what equation(s) the program uses to calculate the range of the projectile with air resistance (air resistance enabled in the program)

2. Jul 1, 2005

### Dr.Brain

You just need to know the direction(s) of the resistance force.Assuming the projectile takes place in x-y plane and the air resistance force is given by:

$F_r=F_x i + F_y j$

You need not remember the formulae, just apply newton 'kinematic equations' in x direction and y direction seperately , in this case taking into consideration the 'acceleration/decelleration due to air resistance force'

BJ

3. Jul 1, 2005

### pete worthington

Air resistance does not fit into a simple equation. Force of air resistance is not a constant, it changes as a function of projectile velocity. You can assume

F drag = C * v or F drag = C * v^2

C is the coefficient of drag for an object. It depends on numerous variables.
Experiments show that the later relationship is a better curve fit. You can bet that the equation(s) they used are formulated on a combination of theoretical and experimental results. You might consider going to the source and asking them.

4. Jul 2, 2005

### SGT

As Pete explained, air resistance is a nonlinear function of velocity, so it is time varying. The only way to solve the differential equations of the movement is numerically. You attribute an initial value to the velocity, calculate the drag and from that the accelerations in the horizontal and vertical directions. You give a small time increment and calculate the new value of the velocity. Work iteratively until you find that the height is zero. The horizontal distance at that instant is the range.

5. May 23, 2008

### jimvoit

The model (differential equation) of the projectile motion will depend on the relationship between the resistive force and the velocity of the projectile that is chosen. Will it be f=k v, f=k v_squared? Wikipedia has some relevent comments in this area and the site at math.fullerton.edu/mathews/n2003/ProjectileMotionMod/ (continues beyond my page) has Mathematica programs which contain the differential equations for the f=k v case and the f=k v_squared case.

6. May 23, 2008

### Shooting Star

Hi Thaer,

The web-program you've mentioned very likely uses the air resistance as proportional to v or v^2, as has been mentioned by others. For the kv, you get a nice analytical solution in 2-d. Not so in the kv^2 case. But when the object is thrown vertically upward, both equations can be integrated quite easily in 1-d.

All the parameters like height, time etc are shown in the program. From this, after solving the kv or kv^2 case analytically, it will be possible to say which one matches your results when you put 90 deg in the program. Maybe the air resistance is not either of these in the program, but some other power or function of v.

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Hi Dr.Brain,

Don't we all wish that things were just always so simple...

Last edited: May 23, 2008