# 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/s^{2}. 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/s
^{2}. For example, if a pilot in a jet is experiencing an acceleration of 24 m/s^{2}, then he is also experiencing an acceleration of (24 m/s^{2}/ 9.8 m/s^{2}=) 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/s^{2} 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/s^{2}. Using our displacement and velocity formulas

x = x_{o
+} v_{o}t + ½ at^{2} and v = v_{o}+ at

let x_{o} = 0m, v_{o} = 0m/s, and
a = g.

Then, x = .5gt^{2}, or x = - 4.9t^{2}. 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/s^{2}(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/s^{2}.

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:*