Re: ? about projectile motion w/ air resistance

On Mar 4, 9:02 pm, brad2...@xxxxxxxxx wrote:
On Mar 4, 4:45 pm, "Phil Holman" <piholmanc@yourservice> wrote:





<brad2...@xxxxxxxxx> wrote in message

news:fc17c13f-6e56-4b32-bf77-0ab3529f5415@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

I'm trying to model projectile motion with air resistance in MATLAB.
I've got my code working but there is a situation that I just can't
figure out.

There is a situation where if you fire a projectile in a pseudo
straight-line (i.e. a little above the straight line but not a lob)
the projectile won't hit the target. Yet, if you lob the target with a
much larger angle to the horizontal, then you will hit the target. In
other words, sometimes you have to lob the projectile and I'm trying
to understand this case.

If this situation makes sense, then I could use some insight into the
problem.

Your discription isn't really clear. What do you mean by straight line
and in which direction? In what way doesn't the projectile hit the
target and where is the tartget? What do you mean by lob the target?

Phil H

I agree about my description..

Basically, projectile motion is taught in physics 101. But, I'm taking
a differential equations course and we are studying the effects of air
resistance on a differential eqn.  Therefore, the motion is not
parabolic but rather drops right down real fast after hitting the top
of the arc.

So, when I talk about a lob I mean a projectile that is shot up in the
air with a big angle to the horizontal. Therefore, it goes really high
and hits the target on the backside of the curve. When I shoot a
projectile in a shallow angle, it doesn't go too high but goes far. In
this case, the projectile overshoots my target.

So my code has to decide when to lob a projectile verses fire at it
with a more shallow angle.

If a target is close in to the firing position and close to the
ground, you need to lob it in order for the projectile to hit.
Otherwise, you will always overshoot.

Hopefully, this helps and makes it more clear..

Thanks for any insight, help, wisdom.. 8-)

-brad w.- Hide quoted text -

- Show quoted text -

OK, let's start with the case where the launch and the landing is at
the same elevation.
In general, there will be one angle at which the range of the
projectile is maximized. For the case of no friction, this is at 45
degrees. For the case with air friction this angle will generally be
higher than 45 degrees. Call this angle A_max.
For all targets at a distance LESS than this maximum range, there will
be TWO angles that the gun can be fired at, one smaller than A_max and
one higher than A_max, that will hit this target.
If you are not seeing two angles that will hit this target, then
you've got some constraints on the gun angle that are preventing you
from seeing both.

You can see why this is analytically for the case of no air friction,
because when you solve the algebraic equations for the angle A in
terms of the desired range x, you'll get a quadratic equation for A --
2 solutions. Naturally adding in air friction doesn't physically
eliminate one of the solutions magically, though it does change the
numerical answer.

I believe, from your description, that you have a case where the gun
launches at some height *above* the landing point, no?

PD
.

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