Problem 71 - Algebra A Combined Function

Question

Neglecting air resistance, the distance \(s(t)\) in feet traveled by a freely falling object is given by the function \(s(t)=16 t^{2}\), where t is time in seconds. Use this formula to solve Exercises 71 through \(74 .\) Round answers to two decimal places. The Petronas Towers in Kuala Lumpur, completed in 1998, are the tallest buildings in Malaysia. Each tower is 1483 feet tall. How long would it take an object to fall to the ground from the top of one of the towers? (Source: Council on Tall Buildings and Urban Habitat, Lehigh University)

Step-by-Step Solution

Answer

9.63 seconds

Step 1: Identify the Given Information and Equation

The problem provides the formula for the distance an object falls as \(s(t) = 16t^2\), where \(s(t)\) is the distance in feet and \(t\) is the time in seconds. The height from which the object is falling is 1483 feet.

Step 2: Set Up the Equation to Solve for Time

Set \(s(t) = 1483\) to find the time \(t\) it takes for the object to fall. This gives the equation:\[16t^2 = 1483\]

Step 3: Solve for Time \(t\)

Divide both sides of the equation by 16 to isolate \(t^2\):\[t^2 = \frac{1483}{16}\]Calculate \(\frac{1483}{16}\), which equals approximately 92.6875.

Step 4: Find the Square Root to Isolate \(t\)

Take the square root of both sides to solve for \(t\):\[t = \sqrt{92.6875}\]\(t \approx 9.63\) (rounded to two decimal places).

Step 5: Conclude the Calculation

The time it takes for an object to fall from the top of the Petronas Towers is approximately 9.63 seconds.

Key Concepts

gravity

Gravity is an invisible force that pulls objects towards each other. On Earth, it gives weight to physical objects and causes them to fall towards the ground when dropped. The strength of gravity on Earth is approximately 9.8 meters per second squared. However, in many physics problems like the one we are considering, gravitational acceleration is simplified to 32 feet per second squared, which when halved gives the constant 16 used in the distance formula

• The gravitational constant: This equation uses the gravitational constant simplified for objects falling freely in a vacuum—meaning no air resistance or friction.
• Free fall: Describes the motion of an object falling solely under the influence of gravity. This is when objects fall without any other forces acting on them.

Understanding gravity helps us solve the exercise because the equation for distance in free fall relies on this constant force.

quadratic equations

Quadratic equations are polynomial equations of degree two. They are commonly written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. The term "quadratic" comes from "quad" meaning square, because the variable gets squared.

• Structure: In our problem, the quadratic is hidden inside the free-fall equation \(16t^2\). Here, \(16\) acts like the "a" coefficient, and "b" and "c" are zero.
• Application: You focus on solving these equations to find when they equal a specific number, making them essential for predicting outcomes like how long it takes a falling object to reach the ground.

By understanding quadratic equations, you learn how to manipulate and solve equations to find unknown values, like time in our exercise.

square root calculation

Square root calculations are necessary when solving quadratic equations, especially when you need to determine a value like time from the equation \(t^2 = 92.6875\). Taking the square root of both sides is a method used to solve for the variable squared.

• Square Root Basics: The square root of a number is a value that, when multiplied by itself, gives the original number. In symbols, if \(x^2 = y\), then \(x = \sqrt{y}\).
• Application in Exercise: To find how long the object falls, calculate \(t = \sqrt{92.6875}\), which gives approximately 9.63 seconds when rounded to two decimal places.

Using square root calculations, you can solve quadratic equations, helping you determine precise results for problems involving free fall and other quadratic equations.