physics question?

Question

What is the appropriate equation for how fast an object freely falling from a position of rest? For how far the object falls?

Answer 1

You ask two questions.

The velocity of an object falling freely from rest is

v = at

where velocity is in meters/second, t is time in seconds, and a is the acceleration due to gravity, 9.81.

The distance fallen in t seconds is given by

d = (1/2)*a*t^2

where d is the distance in meters.

Answer 2

Let t=time, v=velocity, and d=displacement from the initial position. Then:
v=-9.8t
d=-4.9t²
The negative represents that the pull of gravity is downward - if you use a coordinate system that represents down as the positive direction, you would flip that sign. Note also that one must use the appropriate units for these coefficients to be correct - distance in meters, velocity in meters per second, acceleration in meters per second per second, and time in seconds. Finally, note that this is only a special case of the more general equations
v=at+v_0
d=at²/2 + v_0 t + d_0
where a is a constant acceleration, v_0 is the initial velocity, and d_0 is the initial displacement from the origin. These in turn are special cases of the even more general equations:

v=∂d/∂t
a=∂²d/∂t²

Cheers.

Answer 3

A falling object accelerates at the rate of 32 feet/sec^2 (or about 10 m/sec^2). This means that after one second it is falling at the rate of 32 feet/sec. After 2 seconds, it is falling at the rate of 64 feet/sec. Obviously, some information is missing because we don't know the height that the object was at when it fell.

Answer 4

Acceleration of 9.803 meters per second every second

Answer 5

v2=u2+2as

the v and u are squared.

Answer 6

f(t)=9.803(t)
His answer in equation form...
I dunno...

What the hell is "u"?