basic projectile Physics 101 problem......

[razr]

Just lazy or something else, I guess.

Some people like to interact and hear stories or find out what others are doing rather than looking at formulas and charts. Its a complicated subject and regardless of what you read on WIKI, you still have to go out there and experience it.

[Squid]

tip of gun barrel and target are prefectly level and this is in a vacuum camber with

no air resistance, etc, etc bullet and target are like totally tiny, etc

Like I said "bonehead physics class math problem" not real world, but I'm interested in getting into all that later, like how with fairly simple calculus they take into account how air density changes as artillery shell travels up. IIRC there are about 1/2 dozen 'meaningful' things that need to be accounted for like windage from spin of bullet, air density and how it changes, humidity(diff effect than its effect on density....trick question: which weighs more, a cubic mile of air with 5% humidity or cubic mile of air at 95% humidity, given STP(standard Temp and Press) or any equal temp and press? The dry air weighs more than humid air because in vaporous state a molecule of water acts like a molecule of gas, and a N2 or O2 molecule is heavier than a H20 molecule)

The WW1 German "Paris Gun" got longer range by firing at an angle greater than 45' to get the shell up into thinner air with much less drag. Almost a "ICBM" and 1/2 way to outer space!


anyways....Russ is correct we have all the info we need, and we CAN find the time once we do some voodoo and figure out the relation between x and y and the initial velocity of bullet at its angle and sub in to one equation after isolating "t" in another.

Which I will now ponder. I'm guessing it will be a quadratic because Common Sense says there will be two very different angles, low(normal) and high(mortar).



Maybe being able to answer this question correctly should be what the Govt requires for someone to own a gun. If you can't tell me the most basic info needed to point you gun so it hits what you intend what the heck are you doing with a gun in the first place???

I remember when going to elementary school up in mountains of Colorado where long range hunting was big at the Jr HS the math and science teacher announced he would be teaching these equations and sparked great interest in otherwise math adverse students.

[Squid]

YEMFF, you got it correct! thx, now I just have to wrap my head around it.

I almost had it but messed up my trig algebra somewhere.


RibKick, we are supposed to know how and be able to do this stuff backward and forwards and inside out, not just find equations on WWW and plug in numbers.

[Peter.Steele]

I almost had it but messed up my trig algebra somewhere.


RibKick, we are supposed to know how and be able to do this stuff backward and forwards and inside out, not just find equations on WWW and plug in numbers.

Well, here's the thing, you've still managed to "just find the equation on WWW and plug in numbers." You probably still wound up getting the information from Wikipedia, it's just that you asked the question here and someone else googled it.

Whether or not you understand what you're doing doesn't depend on the source of your info.

[vikingm03]

According to my physics class i took last year this is the answer the OP is looking for

I think you two are assuming the bullet travels at 500m/s in the x direction. This is not true if it has a projectile motion, as its x component of velocity will now be 500*cos(theta). While I haven't checked the numbers, what Yemff posted is the way that you would go about solving this problem.

[Yemff]

Thanks for the vote of confidence, and no I didn't just google the answer, I derived it myself and wasted a bit of work time to do it.I had never done projectile motion problems like this before so hopefully it was a learning experience for more than one of us

[SWalt]

I love this. I'm always amazed at the questions asked when a simple google search will give you your answers in spades.

Just lazy or something else, I guess.

Using Google IS the lazy way. God forbid there will be an EMP event or worse, we would really be back in the dark ages. "Whats (10)(10)? Wheres google or my calculator?!"

The distance, velocity and gravity is all that is needed. You are solving for the angle of elevation. Gravity is constant and the bullet drops the moment it leaves the muzzle. The additional distance traveled in its arc is irrelevant. Solve the right triangle. There is no humidity, air temperature, spin drift, coriolis effect, elevation above mean sea level, ballistic coefficient etc to consider.

[cabinetguy]

regardless of the angle, if I fire the gun, the bullet will miss

[Squid]

OK Yemff, now how to find the 2nd possible angle (mortar high angle)???

common sense tells us there will be a 2nd angle.

if it can be reached at under 45' with zero air resistance there has to be a 2nd angle.

I tried 0.06 - 90' but that wasn't it, unless my arthritic was off.

[thegiff]

[joefreas]

Yes, these assumptions make for great problems in the classroom but no ammount of calculation comes close to live fire plotting.

[theicecreamdan]

Here's what I came up with using live (simulated) fire plotting

\

~ (90-0.8985)

[Squid]

why don't you think a high angle solution also exists?

from wiki....

PS-really awkward to type math equations on regular computer, someone need to fix that....I can never make my division ______ things line up. always a mess, and don't get me started about not being able easily type the extra symbols.......


Angle \theta required to hit coordinate (x,y)
Vacuum trajectory of a projectile for different launch angles

To hit a target at range x and altitude y when fired from (0,0) and with initial speedv the required angle(s) of launch \theta are:

\theta = \arctan{\left(\frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx}\right)}

Each root of the equation corresponds to the two possible launch angles so long as both roots aren't imaginary, in which case the initial speed is not great enough to reach the point (x,y) you have selected. The greatest feature of this formula is that it allows you to find the angle of launch needed without the restriction of y = 0.

Derivation

First, two elementary formulae are called upon relating to projectile motion:

x = v t \cos \theta , t = \frac{x}{v \cos \theta} (1)

y = vt \sin \theta - \frac{1}{2} g t^2 (2)

Solving (1) for t and substituting this expression in (2) gives:

y = x \tan \theta - \frac{gx^2}{2v^2 \cos^2 \theta} (2a)

y = x \tan \theta - \frac{gx^2 \sec^2 \theta}{2v^2} (2b) (Trigonometric identity)

y =x \tan \theta - \frac{gx^2}{2v^2}(1+ \tan^2 \theta) (2c) (Trigonometric identity)

0 = \frac{-gx^2}{2v^2} \tan^2 \theta + x \tan \theta - \frac{gx^2}{2v^2} - y (2d) (Algebra)

Let p = \tan \theta

0 = \frac{-gx^2}{2v^2} p^2 + xp - \frac{gx^2}{2v^2} - y (2e) (Substitution)

p = {\frac{-x\pm\sqrt{x^2-4(\frac{-gx^2}{2v^2})(\frac{-gx^2}{2v^2}-y)}}{2(\frac{-gx^2}{2v^2}) }} (2f) (Quadratic formula)

p = \frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx} (2f) (Algebra)

\tan \theta = \frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx} (2g) (Substitution)

\theta = \tan^{-1}{\left(\frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx}\right)} (2h) (Algebra)

Also, if instead of a coordinate (x,y) you're interested in hitting a target at distance r and angle of elevation \phi (polar coordinates), use the relationships x = r \cos \phi and y = r \sin \phi and substitute to get:

\theta = \tan^{-1}{\left(\frac{v^2\pm\sqrt{v^4-g(gr^2\cos^2\phi+2v^2r\sin\phi )}}{gr\cos\phi}\right)}