Problem 64 - A Graphical Approach to Precalculus with Limits

Question

An astronaut on the moon throws a baseball upward. The astronaut is 6 feet, 6 inches tall and the initial velocity of the ball is 30 feet per second. The height of the ball is approximated by the function $$s(t)=-2.7 t^{2}+30 t+6.5$$ where \(t\) is the number of seconds after the ball was thrown. (a) After how many seconds is the ball 12 feet above the moon's surface? (b) How many seconds after it is thrown will the ball return to the surface? (c) The ball will never reach a height of 100 feet. How can this be determined analytically?

Step-by-Step Solution

Answer

(a) 0.18s and 10.18s; (b) 11.70s; (c) Max height is 89.89 feet.

Step 1: Convert Measurements

First, convert the astronaut's height to a uniform unit. Since 6 feet 6 inches equals 6.5 feet, we'll be working in feet without further conversion needed.

Step 2: Set Up the Quadratic Equation for 12 Feet

To find when the ball is 12 feet above the moon's surface, set the height function equal to 12: \(-2.7t^2 + 30t + 6.5 = 12\).

Step 3: Simplify the Equation

Subtract 12 from both sides to set the equation to zero: \(-2.7t^2 + 30t + 6.5 - 12 = 0\), which simplifies to \(-2.7t^2 + 30t - 5.5 = 0\).

Step 4: Solve for t (Time) for Part a

Use the quadratic formula where \(a = -2.7\), \(b = 30\), and \(c = -5.5\): \[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. Calculate \(b^2 - 4ac = 900 + 59.4\), giving \(959.4\).Compute the solutions for \(t\) using the formula.

Step 5: Implement Quadratic Formula

Calculate the time using the quadratic formula: \(t = \frac{-30 \pm \sqrt{959.4}}{-5.4}\). Find \(\sqrt{959.4} \approx 30.98\). So: \(t = \frac{-30 \pm 30.98}{-5.4}\). Solve for both possible values of \(t\), and use the positive root for the real time.

Step 6: Summarize Time Solution Part (a)

The ball reaches 12 feet at about \(t \approx 0.18\) seconds on the ascent and \(t \approx 10.18\) seconds on the descent.

Step 7: Solve for Time the Ball Returns to Surface

Set height function to zero: \(-2.7t^2 + 30t + 6.5 = 0\). Repeat solving using quadratic formula.

Step 8: Compute Return Time Using Quadratic Formula

Using quadratic formula and parameters \(a = -2.7\), \(b = 30\), \(c = 6.5\), compute \(t = \frac{-30 \pm \sqrt{900 - 70.2}}{-5.4}\).Since \(\sqrt{829.8} \approx 28.80\), solve \(t = \frac{-30 \pm 28.80}{-5.4}\). Use the positive solution.

Step 9: Summarize Time Solution Part (b)

The ball returns to the moon’s surface roughly at \(t \approx 11.70\) seconds.

Step 10: Analyze for Maximum Height for 100 Feet

To confirm the ball won't reach 100 feet, check the vertex of the quadratic function. The vertex occurs at \(t = \frac{-b}{2a} = \frac{-30}{2(-2.7)} = 5.56\).

Step 11: Evaluate Maximum Height

Compute the maximum height at \(t = 5.56\) using the height function: \(-2.7(5.56)^2 + 30(5.56) + 6.5 \).Simplify to find peak height is approximately 89.89 feet.

Step 12: Conclude on Height of 100 Feet

Since the maximum height is less than 100 feet, the ball will not reach 100 feet.

Key Concepts

Quadratic Equations

In projectile motion, the path of a thrown object can be described using a quadratic equation. This type of equation is generally written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. For this particular problem, our equation is \(-2.7t^2 + 30t + 6.5 = 0\). These equations help us model real-world scenarios, such as the motion of a ball thrown upwards. Quadratics can be used to find various points in the ball's trajectory, like when it reaches a specific height or when it returns to the ground.

The properties of quadratic equations include the parabola's shape, which in this case opens downwards because the coefficient of the \(t^2\) term is negative. The solutions to these equations can be found using the quadratic formula, roots, or factoring, depending on the scenario.

• The vertex represents the maximum height the ball can reach.
• The roots denote the time intervals when the ball is at ground level.

Understanding these elements is crucial for determining critical moments during the ball's flight, such as when it takes off and when it lands back on the moon's surface.

Initial Velocity

The initial velocity of an object plays a crucial role in determining its motion path. In this exercise, the initial velocity is 30 feet per second. This means that at the moment the ball is thrown, it starts with a speed of 30 feet per second upwards.

Initial velocity impacts how high and how far the ball travels. It is a fundamental component of the quadratic function defining the ball's height over time. The greater the initial velocity, the higher and longer the ball will travel before gravity slows it down and brings it back down to the surface.

Because of the moon's lower gravitational force compared to Earth, the dynamics of motion are altered, making the ball travel a different trajectory than it would on Earth. Initial velocity, along with the gravitational pull coefficient (here, \(-2.7\)), depicts the object's peak height and the range it covers in the air. By effectively manipulating initial velocity, we can predict and determine specific points in the motion path of a projectile.

Maxima and Minima

Maxima and minima in a quadratic equation refer to the highest or lowest point on the graph of the equation. In our scenario, since the quadratic curve noticeably opens downwards (indicated by the negative coefficient of \(t^2\)), it features a maximum point. This point signifies the peak height that the ball reaches during its flight.

In mathematical terms, finding maxima or minima is about determining the vertex of the parabola. The parabola reaches its maximum point at \(t = \frac{-b}{2a}\). For this problem, substituting the values \(b = 30\) and \(a = -2.7\) gives \(t \approx 5.56\) seconds for the maximum height. At this time, the ball is at its highest point in the air before gravity pulls it back down.

Recognizing these points aids in analyzing projectile motion effectively:

• It helps in estimating whether certain heights are reachable.
• It assists in calculating the time taken to reach those heights.

These calculations enable a more profound understanding of the object's behavior in the sky.

Vertex of a Parabola

The vertex of a parabola in a quadratic equation is an essential concept in determining the object's maximum height during projectile motion. For the parabolic path \(-2.7t^2 + 30t + 6.5\), the vertex provides crucial insights about the peak point.

The coordinates of the vertex \((h, k)\) can be found using \(h = \frac{-b}{2a}\) and then substituting back to find the height \(k = s(h)\). In our example, \(h\) occurs when \(t = 5.56\) seconds, which is when the ball reaches its maximum height. Substituting back into the original equation provides the height which is approximately 89.89 feet.

This aspect of vertex calculations clarifies why the ball cannot reach 100 feet as the maximum height achieved is below this limit. Furthermore:

• The vertex helps determine the symmetry of the projectile's motion.
• It gives the peak time and height, offering insights into the motion dynamics.

Understanding how to determine and interpret the vertex allows for a thorough comprehension of any parabolic projectile's motion path.