Acceleration of Gravity

In response to a question about the acceleration of gravity:

We dropped marbles from above our heads and did the 8-count to estimate the time of fall to the floor. 

One typical result:  The ball required .5 seconds to fall 2 meters from rest.

From this information we easily reason out the acceleration.  Knowing `ds and `dt we find vAve, then from v0 and vAve we find vf; from v0 and vf we get `dv and from `dv and `dt we get a.  The results are as follows:

Average velocity is 2 m / (.5 sec) = 4 m/s (based on the definition of average velocity)

Initial velocity is zero, and assuming uniform acceleration we conclude that final velocity is 8 m/s (based on the linearity of the v vs. t graph when acceleration is uniform).

Change in velocity is (8 m/s - 0 m/s) = 8 m/s, so acceleration is 8 m/s / (.5 s) = 16 m/s^2.

Alternatively we can write down the four equations of uniformly accelerated motion, circle the symbols for the know quantities, and solve the the unknown quantity in any equation containing the three known quantities.

The third equation can be solved for a.

Starting with

`ds = v0 `dt + 1/2 a `dt^2, subtract v0 `dt from both sides to get

1/2 a `dt^2 = `ds - v0 `dt, then multiply both sides by 2 to get

a `dt^2 = 2 ( `ds - v0 `dt), and finally divide both sides by `dt^2 to get

a = (2 `ds - v0 `dt) / `dt^2.  Plug in known values:

a = (2 * 2 m - 0 * (.5 sec) ) / (.5 sec)^2 = 4 m / (.5 sec^2) = 16 m/s^2.

More accurate experiments show that the acceleration of gravity in the vicinity of the surface of the Earth, at sea level, is 9.8 m/s^2.

Near the surface of the Earth, any freely falling object accelerates downward at about 9.8 m/s^2.

For nonprecise calculations we can approximate this as 10 m/s^2.  If necessary we can later compensate for the 2% approximation error.