Problem 80 - Elementary Algebra

Question

The height of a projectile shot straight up into the air at 80 feet/second from the ground is given by \(h(t)=-16 t 2+80 t\). At what time will the projectile reach 95 feet? Part D: Discussion Board

Step-by-Step Solution

Answer

The projectile reaches 95 feet at approximately 1.57 seconds and 3.06 seconds.

Step 1: Understanding the Problem

We are given the equation for the height of a projectile: \(h(t) = -16t^2 + 80t\). We need to find the time \(t\) when the projectile reaches a height of 95 feet.

Step 2: Setting Up the Equation

To find the time \(t\) when the projectile is at 95 feet, we set the height equation equal to 95: \(-16t^2 + 80t = 95\).

Step 3: Rearranging the Equation

Rearrange the equation into standard quadratic form: \(-16t^2 + 80t - 95 = 0\).

Step 4: Using the Quadratic Formula

We use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -16\), \(b = 80\), and \(c = -95\).

Step 5: Calculating the Discriminant

Calculate the discriminant: \(b^2 - 4ac = 80^2 - 4(-16)(-95)\). First, \(b^2 = 6400\), and \(4 \cdot 16 \cdot 95 = 6080\). Thus, the discriminant is \(6400 - 6080 = 320\).

Step 6: Finding the Roots

Substitute into the quadratic formula: \(t = \frac{-80 \pm \sqrt{320}}{-32}\). Simplify \(\sqrt{320} = 8\sqrt{5}\). Thus, \(t = \frac{-80 \pm 8\sqrt{5}}{-32}\).

Step 7: Simplifying the Solutions

Simplify the solutions: \(t = \frac{80 \pm 8\sqrt{5}}{32}\). Split it into two solutions: \(t_1 = \frac{80 + 8\sqrt{5}}{32}\) and \(t_2 = \frac{80 - 8\sqrt{5}}{32}\).

Step 8: Approximating the Solutions

Calculate the approximate values: \(t_1 = \frac{80 + 17.89}{32} \approx 3.06\) seconds and \(t_2 = \frac{80 - 17.89}{32} \approx 1.57\) seconds. Both times when the projectile reaches 95 feet due to symmetry.

Key Concepts

Projectile Motion

Projectile motion is a fascinating concept in physics that describes the motion of an object thrown into the air. When dealing with projectile motion, the key is to understand how different variables such as gravity, initial velocity, and time interact. In this context, the projectile is launched straight up with an initial speed of 80 feet per second.
The equation given, \(h(t) = -16t^2 + 80t\), describes the height \(h\) in feet as a function of time \(t\) in seconds. Here,

• \(-16\) is the coefficient related to gravity, which affects how quickly the object slows down and reverses direction.
• The term \(80t\) represents the initial velocity.

The projectile reaches its maximum height and then starts to fall back to the ground, creating a symmetrical parabolic path.

Discriminant

The discriminant is a crucial component of solving quadratic equations using the quadratic formula. It provides valuable insights into the nature of the roots. For any quadratic equation of the form \(ax^2 + bx + c = 0\), the discriminant \(D\) is calculated as \(b^2 - 4ac\).
In our specific scenario, the discriminant helps determine how many times the projectile reaches 95 feet. A positive discriminant indicates two real and distinct solutions, suggesting two times at which the height is achieved. A discriminant of zero would mean it only reaches it once, and a negative discriminant would indicate it never reaches this height. In this case, the discriminant is 320, resulting in two valid solutions that the projectile reaches the specific height of 95 feet.

Quadratic Formula

The quadratic formula is a versatile tool used to find the roots of any quadratic equation. It is expressed as \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This formula allows us to directly solve quadratics without factoring or graphing.
In our problem, with the equation rearranged into the standard form, the quadratic formula becomes our go-to for finding the value of \(t\) such that \(-16t^2 + 80t - 95 = 0\).
Utilizing the formula involves:

• Identifying \(a=-16\), \(b=80\), and \(c=-95\).
• Calculating the discriminant \(b^2 - 4ac\).
• Substituting these into the formula to find \(t\).

This method provides a clear path to determine when the projectile is at a specified height.

Solving Quadratics

Solving quadratic equations is a fundamental skill in algebra, necessary for analyzing scenarios like projectile motion. There are several methods to solve quadratics, including factoring, completing the square, and using the quadratic formula.
For this problem, using the quadratic formula was the most effective approach. The equation was rearranged into quadratic form \(-16t^2 + 80t - 95 = 0\), making it ready for applying the formula. With the calculated discriminant, the substitution into the quadratic formula provided two solutions for \(t\). These solutions, once simplified and approximated, gave \(t_1 \approx 3.06\) and \(t_2 \approx 1.57\).
This process demonstrates how quadratic equations can model real-world scenarios and can be solved systematically to find precise answers.