<P> The equations ignore air resistance, which has a dramatic effect on objects falling an appreciable distance in air, causing them to quickly approach a terminal velocity . The effect of air resistance varies enormously depending on the size and geometry of the falling object--for example, the equations are hopelessly wrong for a feather, which has a low mass but offers a large resistance to the air . (In the absence of an atmosphere all objects fall at the same rate, as astronaut David Scott demonstrated by dropping a hammer and a feather on the surface of the Moon .) </P> <P> The equations also ignore the rotation of the Earth, failing to describe the Coriolis effect for example . Nevertheless, they are usually accurate enough for dense and compact objects falling over heights not exceeding the tallest man - made structures . </P> <P> Near the surface of the Earth, the acceleration due to gravity g = 9.81 m / s (meters per second squared; which might be thought of as "meters per second, per second", or 32.2 ft / s as "feet per second per second") approximately . For other planets, multiply g by the appropriate scaling factor . A coherent set of units for g, d, t and v is essential . Assuming SI units, g is measured in meters per second squared, so d must be measured in meters, t in seconds and v in meters per second . </P> <P> In all cases, the body is assumed to start from rest, and air resistance is neglected . Generally, in Earth's atmosphere, all results below will therefore be quite inaccurate after only 5 seconds of fall (at which time an object's velocity will be a little less than the vacuum value of 49 m / s (9.8 m / s × 5 s) due to air resistance). Air resistance induces a drag force on any body that falls through any atmosphere other than a perfect vacuum, and this drag force increases with velocity until it equals the gravitational force, leaving the object to fall at a constant terminal velocity . </P>

What formula can be used to determine the velocity of an object in free fall