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

Answer:

1) its initial velocity is zero, 2) the downward direction as positive

3) h = 25.66 m

Explanation:

This is a free fall exercise.

1) with falls, its initial velocity is zero and the acceleration is constant throughout the path and is equal to the acceleration due to gravity.

2) a widely used selection to estimate the downward direction as positive

3) We know that for the second part of the fall

         y₀ -y = h/2   at  t = 1 s

        y = y₀ + v₁ t + ½ g t²

where v₁ is the initial velocity of this interval at the point y = h / 2

        v₁ t = (y -y₀) - ½ g t²

        v₁ = h / 2 - ½ g t²

        v₁ = h/2 - g/2

now let's write the equation for the first interval

         v₁² = v₀² + 2 g (y₁ - y₀)

       in this interval v₀ = 0

         v₁² = 2 g (y₁ -y₀)

         v₁² = 2g h/2

we write our system of equations

           v₁² = (h/2 - g/2)²

           v₁² = (2g h / 2)

       

           (h /2 - g/2)² = (2g h / 2)

            h² / 4 - 2  g/2  h/2 + (g/2)² = g h

            h² / 4 - g h/2 - g h + g²/4 = 0

            h² - 3 g h + g² =0              

            h² - 29.4 h +96.04 = 0

we solve the quadratic equation

            h = [29.4 ±√ (29.4² - 4 96.04)] / 2

            h = [29.4 ± 21.91] / 2

            h₁ = 25.66 m

            h₂ = 3.75 m

As the system takes more than 1 S to fall, the correct answer for the height is h = 25.66 m