Problem 204 - Elementary Algebra

Question

In the following exercises, solve. Round answers to the nearest tenth. A stone is thrown vertically upward from a platform that is 20 feet high at a rate of \(160 \mathrm{ft} / \mathrm{sec}\). Use the quadratic equation \(h=-16 t^{2}+160 t+20\) to find how long it will take the stone to reach its maximum height, and then find the maximum height.

Step-by-Step Solution

Answer

It takes 5 seconds to reach the maximum height of 420 feet.

Step 1 - Identify the Given Equation

The given quadratic equation for the height of the stone is: \[ h = -16t^2 + 160t + 20 \] where \( h \) is the height in feet and \( t \) is the time in seconds.

Step 2 - Find the Time at Maximum Height

In a quadratic equation of the form \( ax^2 + bx + c \), the time at which the maximum height is reached can be found using the formula: \[ t = \frac{-b}{2a} \]. Here, \( a = -16 \) and \( b = 160 \). Substitute these values: \[ t = \frac{-160}{2(-16)} \]

Step 3 - Simplify to Find the Time

Solve the equation for \( t \): \[ t = \frac{-160}{-32} = 5 \] So, it takes 5 seconds to reach the maximum height.

Step 4 - Find the Maximum Height

Substitute \( t = 5 \) back into the original height equation to find the maximum height: \[ h = -16(5)^2 + 160(5) + 20 \]

Step 5 - Calculate the Maximum Height

Simplify the equation step-by-step: \[ h = -16(25) + 160(5) + 20 \] \[ h = -400 + 800 + 20 \] \[ h = 420 \] So, the maximum height is 420 feet.

Key Concepts

Vertex Formula

The vertex formula is essential to solving quadratic equations when we are interested in finding the maximum or minimum point of the parabola. For a quadratic equation in the form \[a x^2 + b x + c\]the vertex represents either the highest or lowest point on the graph:

• For \(a > 0\), the vertex is the minimum point.


• For \(a < 0\), the vertex is the maximum point.

The x-value of the vertex can be found using the formula:\[ t = \frac{-b}{2a}\]In our exercise, the quadratic equation is:\[h = -16t^2 + 160t + 20\]Here, \(a = -16\) and \(b = 160\). Substituting these values in the formula gives:\[ t = \frac{-160}{2(-16)} = 5\]This means the stone reaches its maximum height after 5 seconds.

Maximum Height

After finding the time it takes for the stone to reach the maximum height, the next step is to calculate this maximum height. We do this by substituting the time (\(t = 5\) seconds) back into the original equation:
\[ h = -16(5)^2 + 160(5) + 20 \]
Follow these steps inside the equation:

• First, compute \(5^2 = 25\)
• Then, multiply \(-16 \times 25 = -400\)
• Next, compute \(160 \times 5 = 800\)
• Finally, substitute all computed values:
  \[ h = -400 + 800 + 20 \]


• Add them together:
  \[ h = 420 \]

So, the maximum height the stone reaches is 420 feet.

Projectile Motion

Projectile motion refers to the motion of an object thrown into the air, subject to only the acceleration of gravity. Key characteristics include:

• A parabolic trajectory due to the quadratic nature of the motion
• Horizontal and vertical motion components which can be analyzed separately
• The maximum height and the point where the object lands, both calculable using quadratic equations

In our problem, the vertical motion is described by the quadratic function:
\[ h = -16t^2 + 160t + 20 \]
This equation combines constant gravitational acceleration (\(-16 t^2\) ), initial upward velocity (\(160 t\) ), and initial height (20 feet) to describe how the height of the stone changes over time.

Solving Quadratic Equations

Solving quadratic equations is a fundamental skill and involves finding the values of \(x\) (or \(t\) in our exercise) that satisfy the equation. There are different methods to do this:

• **Factoring**: Expressing the quadratic as a product of two binomials when possible.


• **Using the quadratic formula**: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] which provides solutions directly.


• **Completing the square**: Rewriting the quadratic in the form \((x + p)^2 = q\) and then solving for \(x\) .

In our problem, we used the vertex formula (a special case of the quadratic formula) to find the maximum height. This shows the interconnectedness of different methods when working with quadratic equations.