An object thrown directly upward is at a height of h feet after t seconds, where \(\mathrm{h} = -16 (\mathrm{t}...

1. Translate the problem requirements: We need to understand that \(\mathrm{h} = -16(\mathrm{t}-3)^2 + 150\) represents height over time, find when the object reaches maximum height, then calculate the height 2 seconds after that maximum occurs.
2. Identify the maximum height timing: Recognize that this quadratic function in vertex form tells us directly when the maximum occurs.
3. Calculate the target time: Determine the specific time that is 2 seconds after the maximum height is reached.
4. Substitute and evaluate: Plug the target time into the height equation to find the final answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking in plain English. We have an object (like a ball) thrown straight up in the air. The equation \(\mathrm{h} = -16(\mathrm{t}-3)^2 + 150\) tells us the height of this object at any given time.

What we need to find: The height of the object exactly 2 seconds after it reaches its highest point.

Think of it this way: imagine throwing a ball up - it goes up, reaches a peak, then comes back down. We want to know how high the ball is 2 seconds after it hits that peak height.

Process Skill: TRANSLATE - Converting the mathematical language into a concrete, understandable scenario

2. Identify the maximum height timing

Now let's figure out when the object reaches its maximum height. Looking at our equation \(\mathrm{h} = -16(\mathrm{t}-3)^2 + 150\), this is written in what's called "vertex form."

Here's the key insight in plain English: The part \((\mathrm{t}-3)^2\) will always be positive or zero. Since we're subtracting it (the -16 makes it negative), the height h will be largest when \((\mathrm{t}-3)^2 = 0\).

When does \((\mathrm{t}-3)^2\) equal zero? When \(\mathrm{t}-3 = 0\), which means \(\mathrm{t} = 3\).

So the object reaches its maximum height at \(\mathrm{t} = 3\) seconds.

At this maximum point, the height is: \(\mathrm{h} = -16(3-3)^2 + 150 = -16(0) + 150 = 150\) feet.

3. Calculate the target time

Now we need to find the height 2 seconds after the maximum occurs.

We found that maximum height happens at \(\mathrm{t} = 3\) seconds.
So 2 seconds after the maximum means: \(\mathrm{t} = 3 + 2 = 5\) seconds.

This is straightforward arithmetic, but it's important to be clear about what "2 seconds after" means.

4. Substitute and evaluate

Now we substitute \(\mathrm{t} = 5\) into our height equation:

\(\mathrm{h} = -16(\mathrm{t}-3)^2 + 150\)
\(\mathrm{h} = -16(5-3)^2 + 150\)
\(\mathrm{h} = -16(2)^2 + 150\)
\(\mathrm{h} = -16(4) + 150\)
\(\mathrm{h} = -64 + 150\)
\(\mathrm{h} = 86\)

So the object is at a height of 86 feet, 2 seconds after reaching its maximum height.

Final Answer

The height of the object 2 seconds after it reaches its maximum height is 86 feet.

Looking at our answer choices: A. 6, B. 86, C. 134, D. 150, E. 214

Our answer of 86 feet corresponds to choice B, which matches the given correct answer.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding "2 seconds after maximum height"

Students often confuse the timing requirement. They might think the question is asking for the height at \(\mathrm{t} = 2\) seconds (the second time mark) rather than understanding it means 2 seconds AFTER the maximum occurs. This leads them to substitute \(\mathrm{t} = 2\) instead of first finding when maximum occurs (\(\mathrm{t} = 3\)) and then adding 2 seconds to get \(\mathrm{t} = 5\).

2. Not recognizing the vertex form of a parabola

Students may not immediately recognize that \(\mathrm{h} = -16(\mathrm{t}-3)^2 + 150\) is in vertex form, where the vertex occurs when the expression inside the parentheses equals zero. Instead, they might try to expand the equation or use calculus to find the maximum, making the problem unnecessarily complicated and time-consuming.

3. Misinterpreting what "maximum height" means

Some students might think the maximum height value (150 feet) is what they need to work with, rather than understanding that they need to find the TIME when maximum height occurs first, then use that time to determine the target time for calculation.

Errors while executing the approach

1. Arithmetic errors in the final substitution

When calculating \(\mathrm{h} = -16(5-3)^2 + 150\), students commonly make errors such as: calculating \((5-3)^2\) as 4 incorrectly, forgetting the negative sign before 16, or making errors in the final addition/subtraction (-64 + 150).

2. Incorrect time calculation

Even if students understand the concept, they might make simple addition errors when calculating the target time. For example, if they correctly identify that maximum occurs at \(\mathrm{t} = 3\), they might incorrectly calculate \(3 + 2 = 6\) under time pressure, leading to substituting \(\mathrm{t} = 6\) instead of \(\mathrm{t} = 5\).

Errors while selecting the answer

1. Selecting the maximum height value instead of the calculated result

Students might correctly calculate that the maximum height is 150 feet and then mistakenly select answer choice D (150) instead of their calculated result of 86 feet. This happens when they lose track of what the question is actually asking for in the final step.

2. Choosing a time value instead of height value

Since the problem involves both time calculations (finding \(\mathrm{t} = 3\), then \(\mathrm{t} = 5\)) and height calculations, students might confusedly select an answer choice that corresponds to a time value rather than the height value they calculated, especially if they see familiar numbers from their work in the answer choices.