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Free Fall
Free fall by definition means an object that is falling without any outside influence. Therefore, it is necessary to neglect air resistance when talking about falling objects. So, from here on out, our discussions will be based on no air resistance. We do this because it is too complicated to consider air resistance in our equations. Second, we need to see nature in her barest sense in order to gain an understanding of the physical laws that govern her. All objects fall toward the earth because of gravity. We will deal with gravity in more detail in chapter seven. So, for now, we will just know that it exists and it pulls on objects, making them fall. While objects are in free fall, they are in a constant state of acceleration due to what is called the acceleration due to gravity, which is denoted by the letter g. The acceleration due to gravity depends on the size of the planet you are talking about. On the Earth the value of g is:  The kinematic free fall equations can be derived from the original kinematic equations.  Notice the first equation. The only difference from the original is the - gt. Because we are dealing with free fall, we substitute the acceleration due to gravity in for the acceleration part of the equation. I usually make the down direction negative and the up direction positive. So, since gravity acts down, you have to substitute a (-g). The second equation is not so obvious from the original. It is a position equation and is derived from x = vot + 1/2at^2. I will not go into details here, but the free fall equation is expanded to include two positions: the final y position and the initial y position. I use y because free fall is vertical, which is associated with the y-axis of a graph. So, instead of using x, I use y. The only other difference is again instead of a, there is a g, and since g is negative, the whole 1/2gt^2 term has a negative in front of it. These equations are not in their simplest form, but I use them because they can handle any problem dealing with free fall without having to use multiple steps. The following examples will help illustrate this point. Suppose we just wanted to drop an object at rest from a certain height. We can use the two equations to determine the object's speed and distance fallen for any instant of time. Afte one second of free fall the final velocity will be, Notice Vo is zero because we dropped it from rest, and Vf is negative because it is traveling in the downward direction. The distance fallen will be, There are a couple of things of point out. First, the initial y, yo, is set at zero even though the object starts at a definite height. It does not really matter which one you start at zero. Another thing to point out is the - 4.9t^2. Since that term is 1/2*9.8, I just use 4.9 to save time. The following diagram will show to scale the distance fallen by an object for the first 3 seconds.  You can see how the distance increases each second. Another thing to note on the diagram is the speed and distance. Why for the first second does the ball travel 4.9 meters when it has a speed of 9.8 m/s? Give up? Remember the ball started at rest with a speed of 0 m/s. So, even though it is traveling at 9.8 m/s after the first second, it had to accelerate up to that speed. Therefore, while the ball is reaching that speed, it has traveld only 4.9 meters. Here is another question. Would a heavier object fall faster than a lighter object? The answer is absolutely not! Weight has nothing to do with g. Every object experiences the same acceleration due to gravity. Well, you might be saying what about a piece of paper. If I dropped a piece of paper next to a book, the book would win, right? That is true. But, put the piece of paper on top of the book and then drop them. You will see that the paper stays with the book the entire way down. What was different? You eliminated the air resistance on the paper. Try it. It is a cool trick. Now let's consider throwing the ball straight up into the air with an intial speed. Suppose the initial speed is 29.4 m/s. What would be the velocity of the ball one second later? Since we know the acceleration is a negative 9.8 m/s^2, and the object is traveling up in the positive direction, it is slowing down 9.8 m/s every second. So, after one second, the object will be traveling 9.8 m/s slower, which in this example would be 19.6 m/s. The equation would look like, After two seconds, the velocity would be + 9.8 m/s. After three seconds, the velocity will be zero! That's right, an object is actually at rest the instant it hits its maximum height. The picture below illustrates this point.  You may have noticed that this picture is exactly identical to the one above, only this time the guy is below the object. The reason: it is identical. There is a nice symmetry between throwing an object up and dropping an object from rest. You might be wondering if the distance traveled is the same. Well, the distance traveled from the dropped object after three seconds was 44.1 m. Let's use the position equation to find the distance of the object thrown up. Since the initial v is + 29.4 m/s, and the time is 3 seconds to the top, the equation becomes, This does indeed equal 44.1 meters! Pretty cool huh? Do you think when the ball falls back down that it will look identical to the ball that was dropped from rest? Why wouldn't it? Since they are both starting from rest, it will be identical. The only difference is the direction.  The cool thing about this is any object thrown into the air will return to the same position with the same speed it was thrown with. Again, this is true if the height is small enough to not allow air resistance to make a big difference. This diagram has some additional items. I have included the distance coverd by the ball on the way up, on the way down, and the total distance which includes the entire trip. Because the ball travels 44.1 meters on the way up and down, the overall distance is 88.2 meters. I have also included the time at each point of the trip so that we can use the equations to illustrate how easy it is to calculate velocity and displacement. Let's use the equations to see what the velocity of the ball will be after 4 seconds. The diagram shows - 9.8 m/s. The equation to use would look like this. It indeed gives us the correct speed. Let's now determine the displacement during those 4 seconds. Why didn't that equal the total distance traveled? Remember, this is a displacement equation. After 4 seconds, the ball is 39.2 meters from its original position. Which means, if we wanted to find the displacement after 6 seconds, the answer better be zero. This is useful to know because we can determine how long it would take an object to come back down if we are only given the initial speed. For example, if an object is thrown up with a speed of 20 m/s, how long will it take to come back down to it original position? The equation would become, Which reduces to, Solving for t yields a time of 4.1 seconds. Air Resistance One thing to consider at this point is without air resistance, what would keep an object from accelerating indefinitely? (Well, until it hit the groud). In a word, nothing. Imagine how fast a rain drop would be traveling if it fell from 5000 feet. It would be going around 390 miles an hour! It would be lethal to go outside during a rainstorm! So, air resistance is a good thing. It causes objects to stop accelerating at a certain speed called the terminal speed or velocity. But, I said we would not consider air resistance any more so I will shut up now. |