# Projectile Motion With Drag [Aprox's are Welcome]

1. Dec 20, 2007

### Ikeness

The problem is to give a projectile an initial velocity such that it lands a certain height and distance away (2D system, only x and y dimensions). The forces to account for are gravity and air drag.
The knowns are

x is target horizontal distance to travel
y is target height to travel
g is gravity
Any drag related constants can be found through experimentation I guess.

http://www.pha.jhu.edu/~broholm/l5/node3.html [Broken]
This web page almost got me started. It has the equations
x = vi_x*t
y = vi_y*t + g*t^2/2

vi, the initial velocity, is what I am solving for. I can solve the above for vi, obviously, but the above does not account for any air drag.

If you could tell me how to put air drag into the above 2 equations, that would be great.

This is for a computer program, so if there is a solution with some approximations I can probably work them in there.

Thanks,
-Joel

Last edited by a moderator: May 3, 2017
2. Dec 21, 2007

### bsimmo

Well both should be considered independent to start with, for a projectile.
Vertical distance (y) is just the effect of gravity on the object. air drag (or resistive friction is just another force.
In the vertical direction it will just counter gravity and you will need to know work out if it is constant force or an accelerating force (i.e. get larger as the object goes faster).

Similar for horizontal distance, but that is based on the time of flight. i.e. (time it takes for the ball to go up and down, the vertical equation).

I would start by assuming for the vertical that 'drag' is averaged out (it will experience air resistance going up and down, assuming you launches/land on the same height)
Horizonatal drag will be an opposing force -ve velocity.

So y does not change an and x probably becomes x=(vInital_x-vDrag_x)*t
So if the drag is equal to the velocity is just goes up and down.

Thinking of examples
An old game called scorched earth is a good axample to look at, it has now been developed into Scorched3D http://www.scorched3d.co.uk/ you should have a look at the code ;)

Note, this assumes drag does not chance (wind velocity is constant, object does not spin etc..)

3. Dec 21, 2007

### Shooting Star

> If you could tell me how to put air drag into the above 2 equations, that would be great.

The air drag is, in general, not a simple function of speed. It can vary directly as v at low speeds to v^4 for high speeds.

One case which is analytically solvable is when the drag force is directly proportional to the speed, and obviously acts opposite to the direction of motion. Suppose the drag force is –bv. Then we can write the eqn of motion as,

md^2v/dt^2 = mg – bv. (The bold letters mean vectors.) The solution is,

v = exp(-bt/m)(g + vi).

Now you can separate into x and y directions, e.g., v = vxi + vyj, g=-gi etc.

This could work as a start for you.