Problem 45 - Beginning and Intermediate Algebra

Question

Solve each problem. When appropriate, round answers to the nearest tenth. A ball is projected upward from the ground. Its distance in feet from the ground in \(t\) seconds is given by $$ s(t)=-16 t^{2}+128 t $$ At what times will the ball be \(213 \mathrm{ft}\) from the ground?

Step-by-Step Solution

Answer

The ball will be 213 feet from the ground at approximately 2.4 seconds and 5.6 seconds.

Step 1: Set up the equation

Given the function for distance: \[ s(t)=-16t^{2}+128t \] Set the equation equal to 213 to find the time when the ball is at that height: \[ -16t^{2}+128t = 213 \]

Step 2: Rearrange the equation

Rearrange the equation to standard quadratic form: \[ -16t^{2} + 128t - 213 = 0 \]

Step 3: Solve using the quadratic formula

Use the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] For the equation \[ -16t^{2} + 128t - 213 = 0: \] Let \[ a = -16, b = 128, c = -213 \]

Step 4: Calculate the discriminant

The discriminant (\( \Delta \)) is given by \[ b^{2} - 4ac \]. Substitute the values: \[ 128^{2} - 4(-16)(-213) \] \[ 16384 - 13608 = 2776 \]

Step 5: Apply the quadratic formula

Substitute the discriminant and coefficients into the quadratic formula: \[ t = \frac{-128 \pm \sqrt{2776}}{2(-16)} \] \[ t = \frac{-128 \pm 52.69}{-32} \]

Step 6: Solve for the roots

Calculate the two potential solutions: \[ t_1 = \frac{-128 + 52.69}{-32} \] \[ t_1 = \frac{-75.31}{-32} \] \[ t_1 \approx 2.4 \] \[ t_2 = \frac{-128 - 52.69}{-32} \] \[ t_2 = \frac{-180.69}{-32} \] \[ t_2 \approx 5.6 \]

Key Concepts

quadratic formula

The quadratic formula is a powerful tool for solving any quadratic equation of the form \( ax^2 + bx + c = 0 \). This formula is defined as:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a\), \(b\), and \(c\) are the coefficients of the equation. To solve a quadratic equation, we substitute these coefficients into the formula.
First, identify the coefficients from your equation. In our exercise, we have:

• \(a = -16\)
• \(b = 128\)
• \(c = -213\)

Next, substitute these into the quadratic formula. Remember to carefully calculate the discriminant and each part of the formula step by step. This helps to minimize errors and find the accurate values of \(t\).
Let's move on to understanding the discriminant.

discriminant

The discriminant is a key part of the quadratic formula. It tells us the nature of the roots of the quadratic equation.
The discriminant (\( \Delta \)) is: \[ b^2 - 4ac \]
For our exercise, substitute the values of \(a\), \(b\), and \(c\): \[ 128^2 - 4(-16)(-213) \]
Calculate the squared and product terms: \[ 16384 - 13608 = 2776 \]
The value of the discriminant here is 2776.
The discriminant's value indicates the following about the quadratic roots:

• If \( \Delta > 0\), there are two distinct real roots.
• If \( \Delta = 0\), there is one real root (the roots are the same).
• If \( \Delta < 0\), there are no real roots (the roots are complex numbers).

Since our discriminant is positive, we have two distinct real roots for our equation. Applying the quadratic formula will give us these two root values.
Now, let's focus on solving quadratic equations using the discriminant and quadratic formula.

solving quadratic equations

Solving quadratic equations involves rearranging them into standard form and then applying the quadratic formula.
First, ensure the equation is in the form \( ax^2 + bx + c = 0 \).
In our given problem, we rearranged the equation to: \[ -16t^2 + 128t - 213 = 0 \]
By applying the quadratic formula, we substitute \(a\), \(b\), and \(c\): \[ t = \frac{-128 \pm \sqrt{2776}}{-32} \]
Break it into two parts for the plus and minus scenarios:

• \( t_1 = \frac{-128 + \sqrt{2776}}{-32} \)
• \( t_2 = \frac{-128 - \sqrt{2776}}{-32} \)

Calculate the square root: \[ \sqrt{2776} \approx 52.69 \]
Substitute back into our formula:
Solving for \( t_1 \):
\( t_1 = \frac{-128 + 52.69}{-32} \approx 2.4 \)
Solving for \( t_2 \):
\( t_2 = \frac{-128 - 52.69}{-32} \approx 5.6 \)
Thus, the ball will be at 213 feet from the ground at approximately \( t \approx 2.4 \) seconds and \( t \approx 5.6 \) seconds. This process of solving quadratic equations helps to find these specific points in time.