Problem 39 - Algebra 2

Question

PHYSICS When an object is dropped from the top of a 50 -foot tall building, the object will be \(h\) feet above the ground after \(t\) seconds, where \(\frac{1}{4} \sqrt{50-h}=t .\) How far above the ground will the object be after 1 second?

Step-by-Step Solution

Answer

The object will be 34 feet above the ground after 1 second.

Step 1: Understand the Given Equation

The equation provided is \( \frac{1}{4} \sqrt{50-h} = t \), where \( h \) is the height above the ground in feet, and \( t \) is the time in seconds. Our goal is to find \( h \) when \( t = 1 \).

Step 2: Substitute Known Value

Replace \( t \) with 1 in the equation: \( \frac{1}{4} \sqrt{50-h} = 1 \).

Step 3: Solve for \( \sqrt{50-h} \)

Multiply both sides of the equation by 4 to isolate the square root: \( \sqrt{50-h} = 4 \).

Step 4: Square Both Sides

Square both sides to remove the square root: \( (\sqrt{50-h})^2 = 4^2 \), which simplifies to \( 50-h = 16 \).

Step 5: Solve for \( h \)

Add \( h \) to both sides to get \( 50 = 16 + h \). Then subtract 16 from both sides to get \( h = 34 \).

Step 6: Conclusion

After 1 second, the object is 34 feet above the ground.

Key Concepts

Square Root Equation

A square root equation involves an expression with a square root. In our example, the equation is \( \frac{1}{4} \sqrt{50-h} = t \). Here, \( \sqrt{50-h} \) represents the square root. Solving such equations typically involves isolating the square root component before proceeding to eliminate it by squaring both sides of the equation.

In our physics problem, we first multiplied both sides by 4 to get \( \sqrt{50-h} = 4t \). Next, by squaring both sides, we removed the square root to simplify as much as possible. This manipulation made it easier to solve for the unknown \( h \), which is our height above the ground.

Remember, when dealing with square root equations, always ensure the result is positive since square roots implicitly define non-negative values.

Physics Problem

The given problem involves understanding how physics is used in real-world situations, specifically in the context of motion under gravity.

Dropping an object from a building means it is influenced by gravitational forces, causing its position to change over time. The distance an object falls is related to the time it has been falling, and this relationship often involves quadratic equations due to the constant acceleration of gravity.

In such problems, equations are frequently designed to reflect physical principles—here, the square root aspect reflects the non-linear relationship between height and time. These kinds of equations become crucial when determining an object's position or predicting future states.

Height and Time Relationship

Understanding the height and time relationship is key in motion analysis. In our problem, the height of an object at a certain time is given by \( \frac{1}{4} \sqrt{50-h} = t \). This equation tells us how the object's height above ground changes over time.

As objects fall, they accelerate due to gravity, making the relationship between height and time non-linear. This is why the equation involves a square root: the time \( t \) to fall a certain distance \( h \) does not increase linearly. Instead, it involves a square relationship, illustrating how quickly or slowly the object approaches the ground.

This relationship helps show how long an object remains at certain heights above the ground, crucial for predicting or calculating time of impacts or motions.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to solve for unknowns. In our example, we manipulated \( \frac{1}{4} \sqrt{50-h} = t \) to find \( h \) after substituting for \( t = 1 \).

The steps involved include substituting known values, isolating terms, and eliminating square roots. After substituting \( t = 1 \), we multiplied the equation by 4 to isolate the square root, leading us to the equation \( \sqrt{50-h} = 4 \). Squaring both sides provided \( 50-h = 16 \), and simple rearrangement yielded \( h = 34 \).

This process is vital for solving equations, allowing us to change the form of an equation to reveal the desired solution, showing the importance of methodical procedural steps in algebra.