An object dropped near the surface of the earth experiences an approximatelyOriginally Posted byGreg E

constant acceleration of the minus sign is conventional as

up is normally taken as positive, and so downward accelleration is negative.

So , and if the initial speed is then .

Integrating again gives:

,

where is the initial height of the weight, for convenience we will

take this to be .

Now we are interested in what is happening when the weight has fallen

. This occurs when:

,

or:

.

Putting this back into the equation for vertical velocity:

.

At the earth surface .

We could have short circuited this calculation by knowing that the change

in potential energy when the weight drops through a height is

,

which will equal the kinetic energy of the weight when it has fallen from rest

through a distance :

RonL