If you have a building that’s 50 feet tall in an item launched at an initial velocity of 5.4 ft./s what can you conclude about how the height of the building in the initial velocity of the item launched affect the maximum height in the time it takes to get there if it takes 2.67 seconds to get the object 30 feet up from the 50 foot building.

Question

GPT 3.5 · Answer

We can use the equations of motion to analyze the scenario. The equation that relates the initial velocity (u), the time taken (t), the acceleration (a), and the displacement (s) is given by:

s = ut + (1/2)at²

Given that the building is 50 feet tall and the item is launched at an initial velocity of 5.4 ft./s, we can conclude that the initial displacement (s) is 50 feet and the initial velocity (u) is 5.4 ft./s.

Using the equation, we can find the acceleration (a) when the object reaches a height of 30 feet from the 50 foot building:
30 = (5.4)(2.67) + (1/2)(a)(2.67)²
30 = 14.418 + (0.335a)

Rearranging the equation:
(0.335a) = 30 - 14.418
(0.335a) ≈ 15.582
a ≈ 46.5 ft./s²

Now, we can determine the maximum height reached by the object in the given time. Let's say the maximum height from the ground is h. The displacement (s) at this maximum height will be h - 50 (subtracting initial height from the maximum height). The time taken to reach this maximum height will be half of the total time, which is t/2 = 2.67/2 = 1.335 seconds. The acceleration (a) remains the same.

Using the equation, we can find the maximum height reached by the object:
h - 50 = (5.4)(1.335) + (1/2)(46.5)(1.335)²
h - 50 ≈ 9.087 + (0.835a)

Rearranging the equation:
(0.835a) = h - 50 - 9.087
(0.835a) ≈ h - 59.087
a ≈ (h - 59.087) / 0.835

Now, comparing this acceleration to the previous one:
(a) ≈ 46.5 ft./s²
(a) ≈ (h - 59.087) / 0.835

Equating the two expressions:
46.5 ≈ (h - 59.087) / 0.835

Solving for h:
h - 59.087 ≈ 46.5(0.835)
h - 59.087 ≈ 38.8275
h ≈ 97.9145

Therefore, we can conclude that the maximum height reached by the item launched is approximately 97.91 feet when it takes 2.67 seconds to reach the object 30 feet up from the 50 foot building.