Problem 74
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. Round answers to two decimal places. The Hoover Dam, located on the Colorado River on the border of Nevada and Arizona near Las Vegas, is 725 feet tall. How long would it take an object to fall from the top to the base of the dam? (Source: U.S. Committee on Large Dams of the International Commission on Large Dams)
Step-by-Step Solution
Verified Answer
It takes approximately 6.73 seconds for an object to fall from the top to the base of the Hoover Dam.
1Step 1: Identify Given Values
We are given the height of the Hoover Dam as 725 feet. The formula for the distance a freely falling object travels is \( s(t) = 16t^2 \). We need to find the time \( t \) when \( s(t) = 725 \) feet.
2Step 2: Set Up Equation
We need to solve the equation \( s(t) = 16t^2 = 725 \) for \( t \). This gives us the equation \( 16t^2 = 725 \).
3Step 3: Solve for \( t^2 \)
Divide both sides by 16 to isolate \( t^2 \): \[ t^2 = \frac{725}{16} \] Calculate \( \frac{725}{16} \) which is approximately 45.3125.
4Step 4: Solve for \( t \)
Take the square root of both sides to solve for \( t \): \[ t = \sqrt{45.3125} \] Calculating the square root gives \( t \approx 6.73 \) seconds, rounded to two decimal places.
Key Concepts
Quadratic EquationsPhysics ApplicationsProblem Solving Steps
Quadratic Equations
Quadratic equations are mathematical expressions that can be written in the form of \( ax^2 + bx + c = 0 \). In the case of freely falling objects, these equations appear naturally while analyzing motion under gravity. The trajectory of objects falling freely follows a parabolic path, which can be described by a quadratic equation.
In our exercise, the equation \( s(t) = 16t^2 \) is a quadratic equation where \( s(t) \), the distance traveled, depends on the square of time \( t \). This particular form highlights that as time progresses, the distance increases quadratically, making it easy to determine when the object will reach a specified distance.
In our exercise, the equation \( s(t) = 16t^2 \) is a quadratic equation where \( s(t) \), the distance traveled, depends on the square of time \( t \). This particular form highlights that as time progresses, the distance increases quadratically, making it easy to determine when the object will reach a specified distance.
- Therefore, solving quadratic equations like \( 16t^2 = 725 \) helps in determining specific moments in physical applications.
- The process involves algebraic manipulations like isolating \( t^2 \) and then finding the square root to solve for \( t \), as we did in the given solution steps.
Physics Applications
In physics, analyzing the motion of freely falling objects provides insight into gravitational influence and motion dynamics. Freely falling objects undertake a uniform acceleration due to gravity, resulting in predictable motion that can be calculated using specific formulas.
The formula \( s(t) = 16t^2 \) illustrates this, where 16 represents half of the gravitational acceleration (approximated at 32 feet per second squared). Such applications are useful in various scenarios, from architects designing tall structures to engineers planning the safe drop of materials.
The formula \( s(t) = 16t^2 \) illustrates this, where 16 represents half of the gravitational acceleration (approximated at 32 feet per second squared). Such applications are useful in various scenarios, from architects designing tall structures to engineers planning the safe drop of materials.
- Analyses like these help predict motion outcomes when air resistance is neglected, allowing for straightforward computations.
- The practical understanding of these principles ensures safe and efficient designs in real-world engineering projects.
Problem Solving Steps
Approaching physics problems with clear, step-by-step problem-solving strategies helps to simplify and address complexity in calculations. Here's how you can tackle problems involving quadratic applications like freely falling objects:
Start by identifying the given values - in this problem, it's the height of the Hoover Dam. Next, clearly set up your equation reflecting real-world parameters into mathematical expressions, as shown by \( 16t^2 = 725 \).
Start by identifying the given values - in this problem, it's the height of the Hoover Dam. Next, clearly set up your equation reflecting real-world parameters into mathematical expressions, as shown by \( 16t^2 = 725 \).
- Begin by isolating the variable of interest, \( t^2 \), through basic algebraic operations such as division or multiplication.
- To solve for \( t \), take the square root of both sides. This simplification technique is crucial in dealing with quadratic forms.
Other exercises in this chapter
Problem 74
Factor. $$ x^{4}-1 $$
View solution Problem 74
Solve. $$ x^{3}+x-3 x^{2}-3=0 $$
View solution Problem 75
For each quadratic equation, choose the correct substitution for \(a, b,\) and \(c\) in the standard form \(a x^{2}+b x+c=0 .\) \(x^{2}=-10\) a. \(a=1, b=0, c=-
View solution Problem 75
Solve. $$ x^{-2}-x^{-1}-6=0 $$
View solution