Problem 75 - Brief Applied Calculus

Question

GENERAL: Impact Velocity If a marble is dropped from a height of \(x\) feet, it will hit the ground with velocity \(v(x)=\frac{60}{11} \sqrt{x}\) miles per hour (neglecting air resistance). Use this formula to find the velocity with which a marble will strike the ground if it dropped from the top of the tallest building in the United States, the 1451 -foot Willis Tower in Chicago.

Step-by-Step Solution

Answer

The marble will strike the ground at approximately 207.71 miles per hour.

Step 1: Identify Given Variables

The problem states that a marble is dropped from the top of the Willis Tower, which is 1451 feet tall. We need to substitute this value into the velocity formula.

Step 2: Substitute the Height into the Formula

The velocity formula given is \(v(x) = \frac{60}{11} \sqrt{x}\). Substituting \(x = 1451\) into the formula, we get:\[v(1451) = \frac{60}{11} \sqrt{1451}\]

Step 3: Calculate the Square Root

Calculate \(\sqrt{1451}\). Use a calculator for the approximation: \(\sqrt{1451} \approx 38.08\).

Step 4: Complete the Velocity Calculation

Substitute \(\sqrt{1451} \approx 38.08\) back into the velocity equation:\[v(1451) = \frac{60}{11} \times 38.08\]Calculate this expression to find the velocity.

Step 5: Final Calculation

Perform the final calculation: \[v(1451) = \frac{60 \times 38.08}{11} \approx 207.71\] miles per hour.

Key Concepts

Velocity Formula

The velocity formula is a mathematical expression used to determine how fast an object is moving when it hits the ground. In our exercise, the velocity formula is given by

• \(v(x) = \frac{60}{11} \sqrt{x}\)

This formula tells us the velocity in miles per hour as a function of the height \(x\), in feet, from which the object is dropped. It involves two components:

• The constant \(\frac{60}{11}\) which provides the proportionality factor, facilitating the conversion of height into a velocity measure in miles per hour.
• The square root of the height \(\sqrt{x}\), indicating that velocity is proportional to the square root of the height.

Understanding this relationship is important since it shows the non-linear nature of gravitational acceleration impacts—the velocity doesn't increase linearly with height. Instead, it grows with the square root of the height due to the gravitational forces involved.

Square Root Calculation

Square root calculation is a key part of the velocity formula, specifically illustrated by the term \(\sqrt{x}\). To calculate a square root effectively:

• Identify the number you need the square root for—in this case, it is 1451.
• Use a calculator to perform the square root calculation, as it often involves irrational numbers that cannot be easily calculated by hand.

In the exercise, the square root of 1451 was approximated as

• \(\sqrt{1451} \approx 38.08\)

Calculating the square root is crucial as it directly affects the final velocity value. It's important to ensure you're precise during this step because any error here can significantly alter the ultimate result. Additionally, understanding this process develops your ability to handle more complex problems involving other irrational or tricky values.

Problem Solving Steps

Breaking down a problem into steps makes complex calculations manageable. The solution to our exercise was achieved through the following structured approach:

• Identify Given Variables: Establish what is known—here, it was the height (1451 feet).
• Substitute Known Values into the Formula: Input the known height into the velocity formula.
• Calculate Necessary Components: Find the square root of the height, a step calculated with precision using a calculator.
• Substitute Derived Values: Place the calculated square root back into the velocity formula.
• Final Computation: Solve the complete velocity expression to get the final speed, which was found to be approximately 207.71 mph.

This logical and clear step-by-step process helps to isolate different parts of the calculation. It ensures each portion of the formula is carefully evaluated before moving to the next step, ensuring high accuracy and understanding throughout the exercise. Having a clear procedure not only aids in solving mathematical problems but builds organizational habits useful in many complex tasks.