Question

The superhero Green Lantern steps from the top of a tall building. He falls freely from rest to the ground, falling half the total distance to the ground during the last 1.00 s of his fall (Fig. 2.30). What is the height h of the building?

SOLUTION GUIDE

IDENTIFY and SET UP

1. You’re told that Green Lantern falls freely from rest. What does this imply about his acceleration? About his initial velocity?

2. Choose the direction of the positive y -axis. It’s easiest to make the same choice we used for freely falling objects in Section 2.5.

3. You can divide Green Lantern’s fall into two parts: from the top of the building to the halfway point and from the halfway point to the ground. You know that the second part of the fall lasts 1.00 s. Decide what you would need to know about Green Lantern’s motion at the halfway point in order to solve for the target variable h. Then choose two equations, one for the first part of the fall and one for the second part, that you’ll use together to find an expression for h. (There are several pairs of equations that you could choose.)

EXECUTE

4. Use your two equations to solve for the height h. Heights are always positive numbers, so your answer should be positive.

Answer 1

Green Lantern's fall implies zero initial velocity and acceleration due to gravity. The fall can be divided into two phases, and using the equations h = ½gt² and h = g(t - 1)² while knowing half the distance is covered in the last second, we can calculate the height of the building.

Step-by-step explanation:

When Green Lantern falls freely from rest, it implies that his acceleration is due to gravity (acceleration g), which is approximately 9.8 m/s² downward, and his initial velocity is zero. To solve for the height h of the building, we consider two parts of the fall: from the top to the halfway point and from the halfway point to the ground. Given that he falls half the total distance in the last second, we need to find his velocity at the halfway point to solve for h.

Using the formula h = ½gt² for the first part of the fall, and h = g(t - 1)² for the second part (where t is the total time of the fall), we can set up two equations representing the two halves of the descent. However, since he falls a distance of h/2 in the last 1.00 s, we can derive that the total distance fallen in the time t-1 is 2/3h. This information along with the two equations can be used to solve for the height h of the building.