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Let's think about falling--the motion of a body dropped from the earth's surface into a very deep pit. Let's assume that what we are dropping is dense enough that we don't have to worry about air resistance (or, conversely, lift)--a cannonball, say. And let's assume that the pit is deep enough that we don't have to worry about it hitting anything and coming to a stop (at least not for a while). And assume that our distances are small enough that we don't have to worry about how the strength of gravity varies as we move closer to and further from the center of the earth.

At the moment of its release, the cannonball has no upward or downward velocity at all. One second later, gravity--accelerating it at a rate of g feet per second per second--that is, every second gravity gives it an additional downward velocity of g feet per second, where on the earth's surface g = 32--has given it a downward velocity of 32 feet per second. Let's adopt the convention that upward numbers are positive and downward numbers are negative, so that after one second the cannonball's velocity is -32 feet per second.

What's its average velocity during the first second? It starts with a velocity of 0, it ends with a velocity of -32, so that its average velocity is -16 feet per second. How far has it fallen after one second? Moving at an average velocity of -16 feet/second for 1 second gives us a total distance of -16 feet: it has fallen 16 feet.

How about after two seconds? What's its velocity after two seconds? After two seconds, its velocity is -64 feet per second. What's its average velocity over the first two seconds? It starts the first two seconds with a velocity of zero, and ends them with a velocity of -64 feet/second, so its average velocity is -32 feet/second. How far has it fallen after two seconds? Moving at an average velocity of -32 feet/second for two seconds gives a distance of -32 feet/second x 2 seconds = -64 feet--four times as far as it had fallen in the first second.

And after three seconds? A final velocity of -96 feet/second. An average velocity of -48 feet/second. A distance of -48 feet/second x 3 seconds = -144 feet--nine times as far as it had fallen in the first second.

And after four seconds? An final velocity of -128 feet/second. An average velocity of -64 feet/second. A distance of -64 feet/second x 4 seconds = -256 feet--sixteen times as far as it had fallen in the first second.

Note the pattern: 0 seconds... 0. 1 second... 1. 2 seconds... 4. 3 seconds... 9. 4 seconds... 16. That's the pattern we see if something is equal to something else squared. We can thus write down a rule--a function--for a falling body. If we use y_{t} to stand for the position of the cannonball after t seconds, and if we note that for each of the first four seconds the distance fallen is sixteen times that number of seconds squared, we can write the rule:

y

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

Where does the 16 come from? It is half of 32, the number g that tells us the strength of gravity at the earth's surface. If the strength of gravity were different, the motion of falling bodies would be different. But it would follow a similar pattern. Thus we can write the more general rule:

y

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

What does the motion of a falling body look like on the Moon, where g = 5?

Posted by DeLong at February 23, 2003 07:29 AM | TrackBack

Comments

Velocity is gt. Distance is the integral of velocity with respect to time or 1/2 gt^2. That's "where the 16 comes from."

Posted by: Fred Boness on February 23, 2003 10:52 AMf = ma

f= fg

height = (vi^2 * sin()^2)/2g

range = (vi^2 * 2sin())/g

Velocity as a function of time

Vxf = Vxi +Axt

xf = final velocity

xi = initial velociity

t = time

A = aclleration

f = ma

f= mg

height = (vi^2 * sin()^2)/2g

range = (vi^2 * 2sin())/g

Velocity as a function of time

Vxf = Vxi +Axt

xf = final velocity

xi = initial velociity

t = time

A = aclleration

Um, id and Fred, I'm sure Prof de Long is perfectly familiar with elementary calculus and its application to Newtonian mechanics (the theory of gravity is why Newton developed calculus). But he's writing these puzzles for his 9 yo son, to give him a sense of wonder for the subject.

Posted by: derrida derider on February 23, 2003 01:57 PMUm, id and Fred, I'm sure Prof de Long is perfectly familiar with elementary calculus and its application to Newtonian mechanics (the theory of gravity is why Newton developed calculus). But he's writing these puzzles for his 9 yo son, to give him a sense of wonder for the subject.

Posted by: derrida derider on February 23, 2003 01:58 PMBrad, you would lose points for writing g=32 and g=5 in our intro physics course. For dimensional values, numbers are meaningless without specified units. This is not a trivial observation, and is well worth teaching your children. Think yen versus dollars...

I have a problem with this calculation or amusement. If you drop a body into a deep pit. The gravitational force would immediately start to decrease and eventually reach zero at the centre of the earth. Am I right?

If so it would be interesting to consider the motion of a body in a mineshaft bored right through the Earth. A global pendulum!

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