In order to illustrate the method, consider this problem:  A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side.  How long does it take the package to reach the ground?   (Example 2-12 in your text is similar to this problem. We provide more detail about setting the problem up below.)

We'll follow the advice given previously and select the direction of +x to be up.  Note that the diagram indicates this.  It also indicates the position of x_o and x and the direction of v_o.  The comments in blue are added for explanatory purposes. They would not be needed in a student solution.

Given: xo = 0   x = -80 m     vo = +12 m/s   v = ?     a = -9.8 m/s²	xo is selected to be zero at the initial position of the package. The final position is negative, because it's lower than xo. The initial velocity is positive, because it's in the direction of +x. (The balloon is ascending when the package is released. Hence, the package is initially ascending.) The final velocity isn't known. However, we may not need to know it. The acceleration due to gravity is negative.  Falling objects have increasingly negative velocity.  Rising objects have decreasing positive velocity.  In either case, the acceleration is negative.
Goal:  Find the time, t, for the package to reach the ground.

Solution: Since the acceleration is uniform, we can select an appropriate dvat equation.  We know xo, x, vo, and a, and we need to know t. The dvat with these 5 variables and no others is x = xo + vot + ½at².
	This is a quadratic equation and can be solved using the quadratic formula. Rearrange into standard form and apply the quadratic formula. Substitute numerical values with units and calculate the roots.
We're interested in the positive root for this problem.  Therefore, our answer is t = 5.5 s.  (See Example 2-12 in your text for an interpretation of the negative root.) Checks::  Units check:  The units of the radical reduce to m/s, the same as the velocity term in the numerator. The quotient of m/s in the numerator and m/s² in the denominator is s, as expected. Sign check: We selected the positive root, since time should be positive. Sensibility check:  The package will move a distance of a little more than 80 m as it first goes up and then back down. 5 seconds is a reasonable amount of time to go that far under the action of gravity.

The solution is complete.  Notice some things we didn't do in solving the problem.  We didn't divide it into two parts, one for the upward motion and another for downward.  There's no need to do that when you select one direction for +x and use corresponding signs on the given quantities.  Hence, this method saves time.

Something else we didn't do was use g to represent the acceleration due to gravity.  We know that the textbook used g, but we're trying to avoid confusion.  If you've given the text a careful reading, you saw that g is defined to be positive...always.  That means if you use g in a dvat equation with the direction up defined to be positive, you have to change a (+) sign to a (-) sign.  For example, x = x_o + v_ot + ½at² becomes x = x_o + v_ot - ½gt².  This creates too many possibilities for mistakes. We've seen students write x = x_o + v_ot - ½at², x = x_o + v_ot + ½gt², and g = -9.8 m/s², all of which are wrong.  Just stick with the single equation x = x_o + v_ot + ½at², and you'll avoid mistakes.