Calculus – deriving the equations for simple projectile motion

We have worked several examples of simple projectile motion – meaning that the acceleration of gravity was constant and vertical, and there was no acceleration in the horizontal direction. (In particular, there is no air resistance.)

I simply handed us four equations and used them in a first… second… and third post. I said I would show how to derive them.

It was junior year in high school that I learned the equations for position and speed as a function of a constant acceleration. I didn’t take calculus until I was a college freshman… and at some point I decided that I knew enough calculus to derive the equations I had been told to memorize two years before.

This is about as elementary as it gets in calculus, but when it was all new to me, it was a thrill to see what it could do for me in physics. I will actually derive them in two slightly different ways.

Here we go.

Happenings – 2011 May 28

Yes, this post went out much later than usual on a Saturday – much later.

I was moving right along, assembling notes – actually, I was getting more Internet resources – when I got a call from a friend reminding me that we were to have lunch together.

Okay, get cleaned up and get dressed… spend a couple of hours eating, walking, and talking… then I took a nap when I got home… now it’s late afternoon and I have a post to write….

First, the world didn’t end last weekend. Why am I not surprised?

I’ve continued looking at Fairchild’s “Color Appearance Models” – that’s good, because there’s a comment out here about CIELab that I think I know how to answer now. But my picking up color again came first; it’s just serendipity that it happened just before I needed it.

I’ve also gotten a little further along in the tensorial treatment of stress and strain; I still can’t reconcile Landau and Lifshitz with elementary engineering… but, hey, one way or another I’ll sort it out. But I don’t know how long it will take.

The only thing out of the ordinary this week was a headline that greeted me on Yahoo:

Seismologists Tried for Manslaughter for Not Predicting Earthquake

Simple Projectile Motion 3 – Between roof and ground

Let’s try a much shorter range problem. The ship and the fort in the previous post were shooting at each other from 10 miles apart, and the ship could not return fire for about a 220 yard interval.

Reduce the muzzle velocity to 20 meter/second; change h to -10 meter for the high ground firing at the low ground. With these numbers, this situation is more like two guys throwing rocks at each other; one of them is on the roof of a 2-story building and the other is at street level.

Call them “roof” and “ground” forces.

Here, then, are the two parameters for the roof firing upon the ground force.

Since pictures are quite powerful, let me show you the results up front:

I claim that the roof forces can hit the ground forces 50 m away, while the ground forces have to get within 30 m to return fire.

Happenings – 2011 May 21

After two weekends during which I could not do mathematics – although I did do mathematics during the week last week – I discovered on Thursday that the world was supposed to end today. Damn, I would lose a third weekend!

I did get a kick out of one thing. The U.S. Centers for Disease Control issued a warning about how to prepare for an attack by zombies. Really.

Of course, once you read the sentence that says that preparations are pretty much the same for any disaster, you realize that they are merely taking advantage of the prediction in order to put out their standard advice about disaster preparation.

Mechanics: Simple Projectile Motion – 2 (Fort and Ship)

Here is a projectile problem that fascinated me, and I’ve been meaning to show it to you. It comes from Neville de Mestre, “The Mathematics of Projectiles in Sport”, 1990. This is the second post about simple projectile motion, so you might want to look at the first one.

Here we go.

A fort is on top of a cliff h meters directly above the ocean. Approaching the fort is a ship whose guns have the same muzzle velocity vo as the guns at the fort….

Find over what range the ship can be fired on, from the fort, without being able to effectively return the fire.

If gh is small compared with vo^2 show that this distance is approximately double the height of the cliff.

So, we need to find two distances: max from fort to ship, and max from ship to fort.

Happenings – 2011 May 14

I’m feeling a bit under the gun this morning. I want to relax and chat about mathematics – but a friend is coming in from out of town at about noon, and I really need to clean up the house.

At the very least, let me assure you that a technical post should go out this Monday evening. The draft is out there… the screenshots have been inserted… all that remains is final editing of the narrative, and inserting one link. Even with a guest in my house, I should have time to do that.

Between my two-day trip last weekend and my guest this weekend, I knew I was going to need some time off from work. I worked Monday, and I was off Tuesday through Friday. So why isn’t the house clean? Because I did mathematics instead.

Happenings – 2011 May 7

This will not be a weekend for doing mathematics. All day, both Saturday and Sunday, will be given over to a major event (a long-planned happy one) in the lives of some friends. I had hoped to get a post ready during the week, but I was too busy with preparations.

So, unfortunately, there will be no technical post on Monday May 9.

Have a nice weekend.

Elliptical Orbits – Deriving Kepler’s Equation

Here is that drawing again, showing the eccentric anomaly E and the true anomaly f. What we’ve done so far in this post and in that post is just use Kepler’s equation

M = E – e Sin E

to move between position and time on an elliptical orbit.

Let’s derive the equation using geometry and trigonometry. (Both Conway & Prussing, and Bate, Mueller & White – see my bibliography page – have this derivation.)