Problem 79 - Precalculus : Building Concepts and Connections

Question

The height of a ball after being dropped from a point 100 feet above the ground is given by \(h(t)=-16 t^{2}+100,\) where \(t\) is the time in seconds since the ball was dropped, and \(h(t)\) is in feet. (a) When will the ball be 60 feet above the ground? (b) When will the ball reach the ground? (c) For what values of \(t\) does this problem make sense (from a physical standpoint)?

Step-by-Step Solution

Answer

The ball will be 60 feet above the ground approximately \(\sqrt{2.5}\) seconds after it is dropped. It will hit the ground after approximately \(\sqrt{6.25}\) seconds. The problem makes sense for \(t\) in the interval \([0, \sqrt{6.25}]\).

Step 1: Find when the ball is 60 feet above the ground

To find when the ball is 60 feet above the ground, set \(h(t) = 60\) and solve for \(t\). This yields the equation \(-16t^2 + 100 = 60\). Simplify the equation to get \(-16t^2 + 40 = 0\). Divide through by -16 to get \(t^2 = 2.5\). Taking the square root gives two solutions, \(t = \sqrt{2.5}\) and \(t = -\sqrt{2.5}\). Since negative time doesn't make sense in this context, we discard the negative solution. Thus, the ball is 60 feet above the ground after approximately \(\sqrt{2.5}\) seconds.

Step 2: Determine when the ball hits the ground

The ball hits the ground when \(h(t) = 0\). Solving \(-16t^2 + 100 = 0\) for \(t\) gives \(t^2 = 6.25\). Taking the square root gives \(t = \sqrt{6.25}\) and \(t = -\sqrt{6.25}\). Again, we only keep the positive solution, so the ball hits the ground after approximately \(\sqrt{6.25}\) seconds.

Step 3: Identify the valid values of \(t\)

The values of \(t\) for which the problem makes sense are \(t >= 0\), because time cannot be negative, and \(t\) must be less than or equal to the time it takes for the ball to hit the ground (i.e., \(t <= \sqrt{6.25}\)). So the valid values of \(t\) are \(0 <= t <= \sqrt{6.25}\).

Key Concepts

Projectile Motion

When you drop a ball from a height, its motion through the air is known as projectile motion. Projectile motion follows a curved path due to the influence of gravity pushing the object down while it continues to move forward. In this case, the ball was released from a height of 100 feet,
and the formula given was carefully designed to calculate its height at any given time. This equation, \(h(t) = -16t^2 + 100\), is a typical quadratic equation used to model vertical motion under gravity.

• The term \(-16t^2\) reflects the effect of gravity on the ball's motion—a constant acceleration downward.
• The initial height, 100 feet, adds to explain where the ball starts falling from.
• Here, time \(t\) is measured in seconds.

The formula can tell us the height of the ball at any specific moment. Understanding how gravity affects an object's flight path is essential when studying projectile motion.

Solving Equations

Solving equations helps us find unknown values by balancing both sides of the equation. In this exercise, different points in time when the ball is at specific heights or when it hits the ground were calculated using the given equation \(h(t) = -16t^2 + 100\).
To solve part (a) when the ball is 60 feet from the ground:

• Replace \(h(t)\) with 60 in the equation, resulting in \(-16t^2 + 100 = 60\).
• Simplify to form \(-16t^2 + 40 = 0\).
• Divide by \(-16\) to isolate \(t^2 = 2.5\).
• Finally, take the square root to determine \(t\).

Since time cannot be negative, only the positive square root is significant. Similarly, for part (b), when the ball hits the ground, substitute \(h(t)\) with zero:

• This results in \(-16t^2 + 100 = 0\).
• Solving gives \(t^2 = 6.25\).
• The solution is \(t = \sqrt{6.25}\), discarding the negative option.

It's essential to carefully follow each step, checking work to ensure you don't miss important details.

Real-World Applications

Quadratic equations like the one in this exercise are valuable in real-world applications, especially in scenarios involving motion, such as:

• Predicting the trajectory of projectiles, such as balls in sports or objects in physics experiments.
• Designing various structures or paths that need precise heights at specific times.
• Calculating physics problems involving free-falling objects to ensure safety and accuracy.

These applications illustrate the need for such mathematical tools in solving everyday problems involving time and distance. Understanding and using these equations provides a better grasp of how objects move in response to forces, which can help in designing better tools and solutions in engineering and technology.
Recognizing that this problem shares roots with these types of applications makes the math more relevant and exciting.

Negative Solutions in Context

A significant part of solving quadratic equations in physics problems is understanding the context and interpreting solutions correctly. Negative solutions often arise mathematically when solving such equations; however, they don't always provide useful information in real-world scenarios. For example, during projectile motion like this exercise,

• The time \(t\) represents seconds after the ball was dropped.
• Negative values of \(t\) imply measurements before the event started, which doesn't make sense physically.

This aspect is critical when solving problems: not all calculated solutions are relevant. By correctly interpreting why negative time is impossible here, it reinforces the importance of discerning viable solutions from non-viable ones in applied mathematics. Discarding negative outcomes is a standard practice, ensuring that the solutions remain meaningful and adhere to the laws of nature and physics.