Here is something to try at a sporting event. Show that the maximum height h attained by an object projected into the air, such as a baseball, football, or soccer ball, is approximately given by h ≈ 1.2t² m , where t is the total time of flight for the object in seconds. Assume that the object returns to the same level as that from which it was launched, as in Fig. 3–58. For example, if you count seconds and find that a baseball was in the air for t = 5.0 s , the maximum height attained was h = 1.2 x (5.0)² = 30 m . The fun of this relation is that h can be determined without knowledge of the launch speed v₀ or launch angle θ₀ . Why is that exactly? See Section 3–8.

Verified step by step guidance

1. Step 1: Recognize that the total time of flight (t) is the time it takes for the object to go up and come back down to the same level.
2. Step 2: Use the kinematic equation for vertical motion: y = v_0y * t - (1/2) * g * t^2, where y is the vertical displacement, v_0y is the initial vertical velocity, and g is the acceleration due to gravity.
3. Step 3: At the maximum height (h), the vertical velocity is zero. The time to reach maximum height is t/2. Therefore, v_0y = g * (t/2).
4. Step 4: Substitute v_0y into the kinematic equation: h = v_0y * (t/2) - (1/2) * g * (t/2)^2.
5. Step 5: Simplify the equation to show that h ≈ 1.2 * t^2.