In the previous section on position, velocity, and
acceleration we found
that *motion with constant acceleration* is given by position functions of
the form:

*x*(

*t*) =

*at*

^{2}+

*v*

_{0}

*t*+

*x*

_{0}

*a*is the acceleration (a constant),

*v*

_{0}is the velocity at time

*t*= 0, and

*x*

_{0}is the position at time

*t*= 0. The velocity and acceleration functions for such a position function are given by the equations

*v*(

*t*) =

*at*+

*v*

_{0}and

*a*(

*t*) =

*a*.

### Free Fall

The first application we will discuss is that of objects in free fall. In
general, the acceleration of an object in the earth's gravitational field is not
constant. If the object is far away, it will experience a weaker gravitational
force than if it is close by. Near the surface of the earth, however, the
acceleration due to gravity is approximately constant--and is the same value
regardless of the mass of the object (i.e., in the absence of friction from wind
resistance, a feather and a grand piano fall at exactly the same rate). This is
why we can use our equations for constant acceleration to describe objects in
free fall near the earth's surface. The value of this acceleration is *a* = 9.8
m/s^{2}. From now on, however, we will denote this value by *g*, where *g* is
understood to be the constant 9.8 m/s^{2}. (Notice that this is not valid at
large distances from the surface of the earth: the moon, for instance, does
*not* accelerate towards us at 9.8 m/s^{2}.)

The equations describing an object moving perpendicular to the surface of the earth (i.e. up and down) are now easy to write. If we locate the origin of our coordinates right at the earth's surface, and denote the positive direction as that which points upwards, we find that:

*x*(

*t*) = -

*gt*

^{2}+

*v*

_{0}

*t*+

*x*

_{0}

*downwards,*while the positive position-direction was chosen to be up.

How does this relate to an object in free fall? Well, if you stand at the top
of a tower with height *h* and let go of an object, the initial velocity of the
object is *v*_{0} = 0, while the initial position is *x*_{0} = *h*. Plugging these values
into the above equation we find that the motion of an object falling freely from
a height *h* is given by:

*x*(

*t*) = -

*gt*

^{2}+

*h*

*x*(

*t*) = 0 and solve for

*t*. We find that at

*t*= the object hits the ground (i.e. reaches the position 0).

### Firing a Bullet Directly Upwards

The equation

*x*(

*t*) = -

*gt*

^{2}+

*v*

_{0}

*t*+

*x*

_{0}

*v*

_{0}. Since the initial position of the bullet is approximately

*x*

_{0}= 0, the equation for this motion is given by:

*x*(

*t*) = -

*gt*

^{2}+

*v*

_{0}

*t*

*x*(

*t*) = 0 again and solving for

*t*we find that either

*t*= 0 or

*t*= 2

*v*

_{0}/

*g*. Well,

*t*= 0 is just the time when the bullet

*left*the ground, so the time at which it will come back, falling from above, must be

*t*= 2

*v*

_{0}/

*g*. Using our knowledge from the previous section,

*v*(

*t*) = -

*gt*+

*v*

_{0}. If we plug in

*t*= 2

*v*

_{0}/

*g*, we find that the velocity of the bullet as it comes back down and hits the ground is -

*g*(2

*v*

_{0}/

*g*) +

*v*

_{0}= -

*v*

_{0}. In other words, the bullet is traveling at the same speed it had when it was just fired, only in the opposite direction.