Need Mathmatical Formula

[Bacon46]

I recently became board and decided to blow a microwave to pieces. (The one in my kitchen took a dump) Part of this microwave was launched straight up. It returned to earth 6 seconds later only a few feet from where it was launched.
I would like to know the formula (or a link where I can find the formula) used to determine the vertical distance this object traveled. I searched the internet with no luck
Any help would be greatly appreciated

[Corona]

Acceleration due to gravity = 9.8 meters per second per second.

3 seconds max height attained would be 9.8 times 3 = 29.4 meters.

96 feet? Straight up? Really? Wonder what can you do when you are not "board"?

[Herro]

A good formula for this, IIRC, would be X=Xo+VoT + (at^2)/2

Where X= distance; V= velocity; a= acceleration; and t= time. The small o's designate initial conditions.

Because the flight path is the same each way up and down, you take time and cut it in half to solve for only one part of the trajectory. So 6s becomes 3s up and 3s down. So plug in the variables for going up:

X= 0 + 0*3 + (9.8m/s * (3)^2)/2

Here you'd get ~44m, which is around 144ft; impressive.

Corona I don't think your units check out; acceleration is in meters per second per second, but multiplying by seconds would yield meters per second, not meters.

[megalomania]

Ack! For gods sake I unlocked the vault o' evil books and removed my physx book from cryogenic suspension in liquid helium inside a nullentropy chest :( Lets see here, chapter 6, projectile motion. I see trig... must get safety gear on... I can't reproduce the theta symbol in html code here, so I shall use (theta).

A projectile is an object that moves in two dimensions under the influence of only the gravitational force.

The start of a projectile's motion, be it thrown by hand or shot from a gun, is called the launch, and the angle (theta) of the initial velocity vi above the horizontal is called the launch angle.

vix = vicos(theta)
viy = visin(theta)
where vi is the initial speed

Now then, the hard part... Adding gravity and such... reconfiguring the equation for unknown initial velocity...

x1 = (2v02sin(theta)cos(theta)/g)

Oh for the love of all that is good, a trig identity! Lord save me! Remember that 2sin(theta)cos(theta) = sin(2theta)

x1 = (v02sin(2theta)/g)

we know our x1, we want to solve for v0

v0 = sqrt(gx1/sin(2theta))
where g = 9.8 m/s2
where x1 = "a few feet" (you will have to measure this)
where (theta) is the angle you launched your projectile (you will have to measure this too)

To find the height of your projectile you use:

yf = yi + viy(delta t) - 1/2g(delta t)2
where (delta t) is the change in time, 6 seconds
yi had better be zero and the object had better of fallen on the same horizontal plane or all these equations are invalid! yi is the initial starting position, which I assume to be "the ground." The starting position does not have to be the physical ground, but for the purposes of these equations whatever your initial starting height is must also be the final height.

Now then, all you need to do is supply the angle of your launch and the distance in METERS between the starting point and the ending point. Meters, not feet mind you. Plug in the numbers and you get your Y final, the height of your projectile. I suppose you can fudge your angle a little since that may be hard to determine... should be slightly less than 90 degrees.

[nbk2000]

What, no factoring in of air resistance, both up and down?

Calcuate the ballistic coefficient, probable angle of attacks, etc, and figure out how that effects your results! :p

[megalomania]

Well that shows how smart you are, everyone knows they don't actually teach real physics in college. Air resistance and all that real world crap is ignored. Why do you think they blew up so many rockets at NASA? All those physics classes and the only thing I learned of practical value is it is easier to pull a car out of a ditch by applying a force perpendicularly to a rope instead of pulling on it.

Don't get me started on the lack of real world usefulness of higher education, I tend to ramble and look for someone to punch. All that money only buys you credentials. The best education is the one you give yourself. That's why I have a stack of books sitting next to me so I can learn what I really want to know, and why I keep my physics book locked away in cryogenic suspension never to see the light of day except for the odd factoid or constant to satisfy the pedantic in me.

While I am riled up, can anyone tell me of a real world application that would require you to integrate such equations as tan3(x)sec3(x) ? Hmm, if I could time travel back and kill Newton there might be no physics or calculus to deal with :)

[chemdude1999]

I spent 5 days a week in a 1 hour lecture and a 2 hour lab (once a week) for a year. We met at the unholy collegiate hour of 7:00 AM. I remember jackshit. 10 hours of required physics and I can't remember any of it. I've learned more studying ballastic coefficents and reloading procedures than I ever temporarily learned in those 2 courses.

I was good in upper calculus. I could do triple integrals in spherical cooridinates fairly easily, but what practical use do they serve? Especially ones involving those goddamned trig functions.

Would you need Newtonian physics to travel back in time? Or could you just skip straight to quantum mechanics and temporal theory? A goddamn paradox and no one has even travelled in time yet. :) :p

[Bacon46]

Thanks to everyone that replied.
It's a lot of work (for me anyway) just to estimate the height that a piece of tin traveled, but as Megalomania said “The best education is the one you give yourself.”
I have learned more attempting to satisfy my curiosity about seemingly useless facts such as this than I ever did in school. Part of the reason could be in the late 60s and early 70s I had other things on my mind.
Dividing the time aloft evenly as Corona suggested was my first thought.
The link below is to an image of the projectile at the moment of launch. It appears to be hauling ass. For part of the assent the object was traveling faster than it was at any point in its descent.
I am assuming the ride up took less time than the ride down so I scraped that idea. But since I don’t know the velocity and “Hauling Ass” doesn’t seem to work in any of the equations an approximate distance is all I am going to end up with anyway.
I definitely have enough info to work with.
Thanks again for your help
http://img411.imageshack.us/img411/4771/adiosmicrowave0001ol2.th.jpg (http://img411.imageshack.us/my.php?image=adiosmicrowave0001ol2.jpg)

[night33]

I must say I agree whole heartedly except for one area of study. Engineering will give you real life knowledge because the professors (well most of them) will only teach you stuff that you can actually use. All other faculties are crap at teaching you useful information but I will defend engineering to the death about the usefulness given to you (plus if you are in Canada you will get a kick ass stamp and iron ring).

[chemdude1999]

Bacon, nice explosion. What did you use (if I may ask)?

[Bugger]

Air resistance to motion of a projectile, directed against the direction of motion, varies approximately with the square of velocity (at least in turbulent flow, which is usually the case - and I once worked out a formula to describe the trajectory), and it also varies according a friction form factor which is determined by the geometry and surface roughness of the projectile. The resulting constant of proportionality has to be experimentally determined.

[chemdude1999]

Perhaps next time use a range finder to determine the height by taking a reading of the projectile at its apex and your distance from the launch point to determine the final side of the triangle (i.e., the height). And then plug the numbers into Pythagorean Theorem: a2+b2=c2, where c is the hypotenuse or longest side.

Check out:
http://www.cut-the-knot.org/pythagoras/index.shtml

I think a laser range finder should be able to "lock on" the flying object long enough to snatch a reading. You'll have to be quick and maybe lead it some. The reading would be the hypotenuse (longest side) of the imagined triangle. The solve for the height. Don't forget to take the square root of your result to get the final answer.

[Bacon46]

Bacon, nice explosion. What did you use (if I may ask)?

Thanks!
56 gm of KClO3 Flash in a 2” (5cm)ID x 2.5”(6.35cm) OD x 10”(25.4cm) long Cardboard tube. End plugs where cardboard and Plaster of Paris.

[Bacon46]

Perhaps next time use a range finder to determine the height by taking a reading of the projectile at its apex and your distance from the launch point to determine the final side of the triangle (i.e., the height).
I think a laser range finder should be able to "lock on" the flying object long enough to snatch a reading. You'll have to be quick and maybe lead it some. The reading would be the hypotenuse (longest side) of the imagined triangle. The solve for the height. Don't forget to take the square root of your result to get the final answer.

The first problem with that is I would have to place myself in a potentially dangerous situation to even attempt to track any of the debris / shrapnel. I wasn’t even holding the camera. It was on a tripod and I was approximately 100 yards (91.44 meters) away behind my truck.
The second problem is there where hundreds of pieces of shrapnel of various sizes flying in all directions. I sometimes have a hard time picking a specific bird out of a passing flock when I am hunting and I’m not worried about one of them taking off my head.
The image below is of the larger pieces of the microwave that where recovered. I recovered around 100 smaller but still potentially lethal pieces that are not shown.

http://img340.imageshack.us/img340/9975/imgp3734dy1.th.jpg (http://img340.imageshack.us/my.php?image=imgp3734dy1.jpg)

[chemdude1999]

You are right. It was just an idea. If you have a range finder for other purposes, you could try it with nothing lost. They can range the distance easily, but the hard part would be in "hitting" the target.

The other idea I had would be placing a object of known height that would be expendable next to ground zero. Then if you capture the apex of the launched item and the indexing object in the same frame, you could get a reasonable estimate of the height.

[Bacon46]

Very true Chemdude. As long as the camera angle is correct, even if I didn’t capture the apex I could determine the initial velocity of the projectile eliminating a major unknown.
I feel another “experiment” coming on.:D
I am all out of microwaves but I do have a computer I’m not using!

[nbk2000]

If you have a videocamera on a tripod, finding the height is easy, as long as you can see the projectiles apex on the video.

After the explosion, without moving the camera, rewind the tape and note where the apex is on the camera monitor (not eyepiece).

Then, with a ballon tethered to the ground at the launch site, raise it until the ballon is at the same height on the monitor as the projectiles apex. Recover the ballon and measure the length of tether used. That's the projectiles max height.

This assumes you have a helper to handle the ballon and there's no wind.