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The last problem established that if you dropped a cannonball down a very deep well, the distance it falls (with positive numbers being up, and negative numbers being down) could be described by the *gravitational fall equation*:

y

_{t}= -(0.5)(g)(t^{2})

That is, the vertical position of the cannonball t seconds after you drop it (that's the "y_{t}" part) is equal to minus (that's because it falls--moves downward) 1/2 times the square of the number t of seconds since you dropped it.

Moreover. on earth where the force of gravity g = 32 feet/second/second, this is the same as:

y

_{t}= -16(t^{2})

But usually we are not interested in cannon balls dropped down wells. Usually we are interested in cannonballs fired from cannon. And they have an initial upward velocity--call it v_{y}, "v" for "velocity" and "y" because we are using the variable y to stand for how high above (or far below) our cannonball is from its launching point. What happens to a cannonball that is not dropped, but is instead initially launched with an upward velocity v_{y}--say, 64 feet/second?

Well, if there were no force of gravity, then after one second the cannonball would be 64 feet high, and after two seconds 128 feet, and after three seconds 192 feet, and after t seconds it would be 64t feet above its launching point. In the absence of gravity:

y

_{t}= 64t

Or, more generally (if it were to be launched with some different upward velocity):

y

_{t}= (v_{y})(t)

But what if we have both (a) an initial upward velocity and (b) gravity? It is a somewhat miraculous fact that--if we are considering small enough distances that we can ignore variations in gravity, if the object is dense enough that we can ignore air resistance and lift, if it doesn't run into anything (like the ground), et cetera, et cetera--all we have to do is to combine the calculations for the two different kinds of motions to get:

y

_{t}= (v_{y})(t) - (0.5)(g)(t^{2})

That tells us what the vertical position of the object will be at any positive time t (if it hasn't run into anything yet). After t seconds, the downward pull of gravity will have moved its position downward by - (0.5)(g)(t^{2}). Its upward initial velocity will have carried it up by (v_{y})(t). And the net position of the object will be just the sum of those two components.

Notice that this equation describes perfectly well what happens to our dropped cannonball. What is a dropped cannonball? Just an object with an initial upward velocity v_{y} = 0. And if v_{y} equals zero, the entire first term is always zero--and we are back to our gravitational fall equation. Conversely, note that this equation describes perfectly well what happens if there is no gravity. If there is no gravity, then g = 0, the second term is always zero, and all we are left with is a cannonball moving upward at its constant upward velocity v_{y}.

Now let's plug in some numbers. Suppose we are on earth, where g = 32 feet/second/second. And suppose we throw a cannonball upward with an initial velocity of 16 feet/second. Then our equation becomes:

y

_{t}= 16(t) - 16(t^{2})

At time t = 0 seconds, this equation tells us, the height y_{0} of the cannonball is zero. That's good: we're just throwing it upward at time 0, and it hasn't had time to leave the ground.

Now let's calculate y_{1}, the height of the cannonball at t = 1 second after it is thrown upward. Substituting "1" in for "t"...

y

_{1}= 16(1) - 16(1^{2}) = 16 - 16 = 0 feet

...uh-oh. Better run fast. At the end of the first second the cannonball is at height zero again, falling downward and hitting the ground (or your foot, if you didn't get out of the way). Halfway between, at time t = 1/2 second, the position is:

y

_{(1/2)}= 16(1/2) - 16((1/2)^{2}) = 8 - 16(1/4) = 8 - 4 = 4 feet

That's as high as it gets.

So now let's consider a much faster cannonball: one launched upward at an initial velocity v_{y} of 320 feet/second. We can make a table of its height:

t seconds after launch... | the cannonball is this many feet high... |

0 | 0 |

1 | 304 |

2 | 576 |

3 | 816 |

4 | 1024 |

5 | 1200 |

6 | 1344 |

7 | 1456 |

8 | 1536 |

9 | 1584 |

10 | 1600 |

11 | 1584 |

12 | 1536 |

13 | 1456 |

14 | 1344 |

15 | 1200 |

16 | 1024 |

17 | 816 |

18 | 576 |

19 | 304 |

20 | 0 |

After ten seconds, the cannonball is lost in the sky: 1600 feet up, it (if it is 1/2 a foot in diameter) is a speck less than 1/60 of a degree across in your visual field. After twenty seconds, the cannonball is back--moving downward this time, it hits the earth with a downward velocity of 320 feet/second.

Posted by DeLong at February 23, 2003 05:03 PM | TrackBack

Comments

The next step is to launch the cannonball at an angle and split the muzzle velocity into vertical and horizontal vectors.

That makes it clear that hitting any target downrange can be accomplished two ways: with a high trajectory launch (between 45 and 90 degrees) or with a low trajectory launch (between 0 and 45 degrees). The maximum range launch trajectory being 45 degrees.

More interestingly, for someone downrange (otherwise known as the target), a couple of radar readouts of the cannonball plus some quick computer calculations will almost instantaneously locate the launch position - for counterbattery fire. Our guys in the Persian Gulf War I had great success with this technique . . . . . .

Posted by: Anarchus on February 23, 2003 07:02 PMBut do I dare introduce the sin() and cos() functions to a twelve-year-old?

Posted by: Brad DeLong on February 23, 2003 08:46 PMEven if I don't dare introduce sin() and cos(), perhaps I can use the Pythagorean Theorem? Muzzle velocity as the hypotenuse, and horizontal and vertical components of the velocity vector as the sides of the right triangle?

Posted by: Brad DeLong on February 23, 2003 08:48 PMOf course, in real life applications you have to also factor in the ballistic coefficient of the projectile.

This is, basically, a mathematical calculation of the projectile's aerodynamic characteristcs. Ballistic coefficient will alter how fast a projectile is traveling at any given point down range after the initial acceleration when fired out of the barrel of a gun or even if simply dropped (due to air drag).

Then you also have to consider altitude and barometric readings; if you want to be precise.

Posted by: E. Avedisian on February 23, 2003 11:58 PMAs an side, I recall reading that many people who can calculate kinematics problems in high school still don't really "get" forces. To illustrate, students were asked to draw the forces on an object on its way up after being thrown. Most people drew an arrow pointing up, because the object is moving up.

Associating force with acceleration rather than velocity, and appreciating that acceleration may have a different direction from the current velocity, is apparently difficult to get.

Don't ask me for the reference though -- it was a physics educational magazine from many years ago.

Posted by: Tom Slee on February 24, 2003 06:21 AMMy favorite "force" story involved a Nolan Ryan interview. He was asked if he threw fast in little league, and he said, "no, I didn't throw that fast back then, but I could throw a lot farther than anyone else . . . . . ".

Posted by: Anarchus on February 24, 2003 06:40 AMBrad,

If your twelve year old can argue you to a standstill and figure out the famous Internet Mind reading, surely he can be trusted with the pythagorean proposition?

That is sufficient to illustrate the components of force. It is a fascinating discussion to use ballistics to understand why the towering homerun uses a lot more energy and travels shorter than the sweetly optimized 45 degree shot.

More fascinating is the extension of gravitational pull to calculate escape velocity - in essence, how fast does it have to go so 'what goes up *doesn't* have to come down'?

Posted by: Suresh Krishnamoorthy on February 24, 2003 07:07 AMyou can be travelling at any arbitrary speed to escape the earth's gravity....

of course you need to be under constant thrust of 1g, but there's no "mandatory" velocity... unless you conced that you need to be ballistic...

maybe play with baseballs and ping pong balls, as they're more real (though of course the calcs are not as interesting)

or you can wip a superball.... then you can account for compression, modulus of elasticity, shearing forces, and concept of energy...

liber, a nit: you need to have a thrust marginally over 1g (to overcome friction) or else you would float in mid air, but you are correct - unless you want to be shot into outerspace from a cannon, you can 'escape' at any velocity.

it is strange looking at g and referring to it at 32 feet/sec/sec - I grew up with 9.8 and 6.67 e -11, numbers that are seared into my memory.

Posted by: Suresh Krishnamoorthy on February 24, 2003 09:03 AMIn transferring from one engineering school to another, I went from the 9.81 to 32 as well. It was very confusing. On the whole, I liked metric better.

Posted by: Anarchus on February 24, 2003 09:51 AMYou don't have to get into trig to demonstrate something really neat -- that the downward acceleration is independent of the horizontal velocity. The classic demonstration of this is the story of the novice hunter and the monkey. The novice hunter with his bow and arrow spies a monkey hanging from a branch. Being a novice, he aims his arrow directly at the monkey. When the monkey sees the arrow fired, he drops from the branch. Unfortunately for the monkey, both he and the arrow accelerate downward at the same rate, so he gets hit.

Many freshman physics classes will do this as a lecture demo. Mine did, complete with a cute stuffed monkey.

Posted by: Curt Wilson on February 24, 2003 05:22 PMThe escape velocity from Earth is given by v|esc^2 = 2*(mu)/r, where r = R + h for a circular orbit of altitude h and Earth radius R, and with mu = 403.5033E12 m^3/s^2.

Mu is the planetary gravitational constant; it differs for each planetary body.

For h = 0, r = R = 6378.140 km, and, therefore, the escape velocity at the Earth's surface is 36,923.5 ft/s after appropriate conversions.

This establishes the energy requirement for such a launch (which would be into a parabolic trajectory).

Doug: Hyperbolic, surely?

Posted by: John on December 16, 2003 01:32 AMJohn: Actually, no. The velocity of escape is also known as the parabolic velocity, a special case in which E, the total energy of the body (kinetic plus potential), is zero. A hyperbolic orbit has energy E > 0, and an elliptical orbit has energy E < 0.

Posted by: Doug Rusta on December 25, 2003 06:44 PMPost a comment