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}
where

*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*.

We will now use these equations to solve some physics problems involving motion
in one dimension with constant acceleration.

####
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}
Notice the

- sign that arises because the acceleration due to gravity points

*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*
If we want to know, for instance, how long it takes for the object to reach the
ground, we simply set

*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}
for an object moving up and down near the earth's surface can be used for more
than just describing a falling object. We can also understand what happens to a
bullet fired directly upwards from the surface of the earth at an initial speed

*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*
How fast will the bullet be traveling when it comes back down and hits the
earth? To answer this we must (i) solve for the time at which the bullet will
hit the earth, and (ii) find the velocity function, so that we can evaluate it
at that time. Setting

*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.