Problem 106 - Algebra A Combined Function

Question

An object is dropped from the top of Pittsburgh's USX Towers, which is 841 feet tall. (Source: World Almanac research) The height of the object after seconds is given by the expression \(841-16 t^{2}\). a. Find the height of the object after 2 seconds. b. Find the height of the object after 5 seconds. c. To the nearest whole second, estimate when the object hits the ground. d. Factor \(841-16 t^{2}\).

Step-by-Step Solution

Answer

a. 777 feet; b. 441 feet; c. 7 seconds; d. \((29 + 4t)(29 - 4t)\).

Step 1: Height After 2 Seconds

To find the height after 2 seconds, substitute \( t = 2 \) into the equation \( 841 - 16t^2 \). Calculate: \[ 841 - 16(2)^2 = 841 - 16 \times 4 = 841 - 64 = 777. \] Thus, the height after 2 seconds is 777 feet.

Step 2: Height After 5 Seconds

To determine the height after 5 seconds, substitute \( t = 5 \) in the equation: \[ 841 - 16(5)^2 = 841 - 16 \times 25 = 841 - 400 = 441. \] Therefore, the height after 5 seconds is 441 feet.

Step 3: Time When Object Hits the Ground

To find when the object hits the ground, set the height expression equal to zero: \( 841 - 16t^2 = 0 \). Solve for \( t \): \[ 16t^2 = 841 \] \[ t^2 = \frac{841}{16} \] \[ t = \sqrt{\frac{841}{16}} = \frac{\sqrt{841}}{4} = \frac{29}{4} = 7.25. \] Thus, to the nearest whole second, the object hits the ground at 7 seconds.

Step 4: Factor the Expression

We need to factor \( 841 - 16t^2 \), which is a difference of squares: \( (29)^2 - (4t)^2 \). This can be factored as: \[ (29 + 4t)(29 - 4t). \] Thus, the factored form is \( (29 + 4t)(29 - 4t) \).

Key Concepts

Factoring

Factoring in algebra involves breaking down an expression into a product of simpler expressions, or "factors." This is useful because it can make equations easier to solve or simplify. In the original exercise, you encounter the expression \[841 - 16t^2\] This expression can be factored because it is a difference of squares. To factor it:

• Recognize that 841 is \((29)^2\) and \(-16t^2\) is \((-4t)^2\).
• Apply the difference of squares formula, which states that \(a^2 - b^2 = (a+b)(a-b)\).
• Applying this formula gives: \((29 + 4t)(29 - 4t)\).

Factoring is like reverse multiplication, so understanding and identifying different patterns and forms help in rewriting expressions efficiently.

Physics Applications

Quadratic equations often model real-world phenomena, especially in physics. When objects move under the influence of gravity, their motion can be described by quadratic functions. In our exercise, the height of an object dropped from a tower is modeled by the equation:\[841 - 16t^2\]This equation gives you the height at any time \(t\).

• The constant 841 represents the initial height in feet.
• The coefficient \(-16\) is derived from gravity and shows the rate at which the height decreases over time.

These applications are crucial because knowing how to manipulate and solve these equations can predict physical behaviors. This example exemplifies how mathematics connects to the real world, showing the object’s height reduces as it falls due to gravity.

Difference of Squares

The difference of squares is a special pattern in algebra where an expression is the subtraction of one square number from another. Recognizing this pattern simplifies complex expressions and helps in factoring tasks. In the scenario, the expression:\[841 - 16t^2 \] is a difference of squares:

• 841 is the square of 29.
• 16t² is the square of 4t.

Using the formula \[a^2 - b^2 = (a+b)(a-b)\]we factored the expression to \[(29 + 4t)(29 - 4t)\]. This specific type of polynomial has easily identified elements that when subtracted give a simplified product or factors. Recognizing these squares makes algebra more approachable and is essential for solving equations.

Algebraic Substitution

Substitution is a powerful algebraic technique where you replace a variable with a given value to simplify or solve an equation. In this exercise, you find the object's height at specific times by substituting the variable \(t\) with numerical values.

• For example, substituting \(t = 2\) gives the height after 2 seconds: \(841 - 16(2)^2 = 777\) feet.
• Similarly, with \(t = 5\), you find the height after 5 seconds: \(841 - 16(5)^2 = 441\) feet.

Substitution helps us understand how the variable changes over time. It's like exchanging a placeholder with a specific value, allowing you to make calculations or derive solutions. It brings abstract mathematical expressions into a concrete, useable result.