Falling Objects
Gravity: It's not just a good idea, it's the law! Near the surface of the Earth and in the absence of air resistance, objects in free fall will have a uniform acceleration of about 9.8 m/s2. This is the value of acceleration due to gravity, which we commonly refer to as g.
- Frequently we will refer to acceleration in terms of g. 1 g = 9.8 m/s2. For example, if a pilot in a jet is experiencing an acceleration of 24 m/s2, then he is also experiencing an acceleration of (24 m/s2 / 9.8 m/s2=) 3.5 gs.
We are going to make a bold leap here and completely ignore the effects of air resistance (drag) and make the following statement: All objects fall with the same downward acceleration due to gravity - 9.8 m/s2 I'm making a big deal of this because it is an important point when we start talking about projectile motion and independent movement in the x-axis and the y-axis. It is thought that Galileo was the first to figure this out and write about it accurately, although he was unable to test his hypothesis in a vacuum. How do we make use of this?
Consider 2 Cases:
In case 1, we are at the top of a 100 meter building and we drop a ball off. Let's call "down" negative, with the value of g = -9.8 m/s2. Using our displacement and velocity formulas
x = xo + vot + ½ at2 and v = vo+ at
let xo = 0m, vo = 0m/s, and
a = g.
Then, x = .5gt2, or x = - 4.9t2. Now we can
generate a table of values for position and velocity:
|
Time |
Displacement |
Velocity |
|
0 sec |
0 meters |
0 m/s |
|
1 sec |
-4.9 m |
-9.8 m/s |
|
2 sec |
-19.6 m |
-19.6 m/s |
|
3 sec |
-44.1 m |
-29.4 m/s |
|
4 sec |
-78.4 m |
-39.2 m/s |
|
4.5 sec |
-100 m |
-44.1m/s |
One thing to notice is, while velocity is increasing linearly, the displacement is increasing exponentially as a square of the time elapsed.
In case 2, we will throw a ball up in the air with a constant velocity - say 20 m/s. The acceleration on the ball is always -9.8 m/s2(as long as we ignore air resistance and gravity is the only force acting on the ball). Thus, the ball will rise to a maximum height where the velocity of the ball is 0 m/s, then will drop back towards the thrower's hand until is returns with a velocity of - 20 m/s. The key point here is that the acceleration, as far as we will be concerned in our problem solving, is ALWAYS -9.8 m/s2.
Often it is helpful to see what someone else has to say on the topic. Try http://www.physicsclassroom.com/Class/1DKin/U1L5a.html for more information. Follow all the threads of this lesson.
For Practice Problems, Try: From the University of Oregon, any of these:
Print 