Problem 42 - Precalculus : Building Concepts and Connections

Question

The height of a ball that is thrown directly upward from a point 200 feet above the ground with an initial velocity of 40 feet per second is given by \(h(t)=-16 t^{2}+40 t+200,\) where \(t\) is the amount of time elapsed since the ball was thrown; \(t\) is in seconds and \(h(t)\) is in feet. For what values of \(t\) will the height of the ball be below 100 feet?

Step-by-Step Solution

Answer

The solutions of the equation \(-16t^2 + 40t + 100 = 0\) give the time instants when the height of the ball is below 100 feet. Discard any negative solutions or solutions that exceed the time the ball is in the air. The range between two valid \(t\) values gives the period when the height of the ball remains below 100 feet.

Step 1: Interpret the problem

Given the height equation \(h(t)=-16t^2 +40t +200\), we need to find the values of \(t\) for which the height of the ball is below 100 feet. To find this, we set \(h(t)=100\) and solve for \(t\).

Step 2: Set up the Equation

So the equation will be: \(-16t^2 + 40t + 200=100\). This simplifies to \(-16t^2 + 40t + 100 = 0\). This equation is a quadratic equation that can be solved for \(t\).

Step 3: Solve the Equation

This quadratic equation can be solved using the quadratic formula \(t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\). Substituting the values from the equation, where \(a=-16\), \(b=40\) and \(c=100\), we get the solutions for \(t\).

Step 4: Check the Solutions

Note: We may get two solutions for \(t\) but only the ones falling within the range of the ball being in the air will be valid. We will discard any negative solutions or solutions that exceed the time the ball is in the air.

Step 5: Find the time range for which the height of the ball is below 100 feet

The height of the ball is below 100 feet between the time it first drops below 100 feet and then goes under 100 feet again when coming down. This is the range between the two valid \(t\) values we found in the previous step.

Key Concepts

Parabolic Motion

When a ball is thrown upward, its trajectory follows a specific path known as a parabola. This is a common phenomenon in projectile motion and is characterized by a symmetrical curved path. The equation given in the exercise, \[h(t) = -16t^2 + 40t + 200\]describes this parabolic trajectory. Here, the coefficient of \(-16t^2\) is responsible for the downward curve, representing the gravitational pull that decelerates the upward motion and accelerates the descent of the ball.

• The peak of the parabola is where the ball reaches its maximum height before descending.

Recognizing the shape and behaviour of parabolic motion helps in predicting the ball's height at any given moment, allowing us to solve for when the height is below certain thresholds, like 100 feet in this scenario.

Physics in Mathematics

The exercise intertwines physical concepts with mathematics, specifically linking kinematic equations with algebraic expressions. Physics in mathematics often involves translating equations realized in mathematical form to reflect real-world phenomena. In our problem:

• The mathematical equation \(h(t) = -16t^2 + 40t + 200\)
• Represents the physical scenario of a ball's motion, dictated by forces such as gravity.

In physics, time variable \(t\) is often crucial as it indicates how a system evolves, while the function of height \(h(t)\) shows how the position of the ball changes over time. Understanding this link allows students to see how equations can model physical systems, predict future states, and solve for unknown quantities.

Solving Inequalities

To determine when the height of the ball is below 100 feet, students need to interpret and solve the inequality:\[-16t^2 + 40t + 200 < 100\]This inequality must be solved to discover the range of time values \(t\) where the ball is below 100 feet.

• First, the inequality is rearranged to resemble a standard quadratic expression: \(-16t^2 + 40t + 100 < 0\)
• Then, techniques such as factoring, completing the square, or applying the quadratic formula can be utilized to solve the quadratic expression.

On solving, the solutions indicate the time intervals. Remember to check that these solutions fit the realistic time frame of the ball’s entire motion.

Kinematics

Kinematics is a branch of physics that describes how objects move. In this problem, kinematics helps us understand the motion of the ball, including its position (height) and velocity at any time \(t\).

• The initial velocity is given as 40 feet per second, influencing how fast the ball ascends.
• Gravity, inherent in the \(-16t^2\) term, gradually slows the ascent and speeds the descent.

These elements combine in the given equation to completely describe the motion of the ball. By breaking down the factors affecting the motion, students can better predict when and how the height changes, which is crucial to solving the exercise effectively.

Quadratic Formula

The quadratic formula is a powerful mathematical tool used to solve quadratic equations of the form \(ax^2 + bx + c = 0.\) In our exercise, the equation\(-16t^2 + 40t + 100 = 0\)fits this pattern.The formula is expressed as:\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

• For our equation, the coefficients are \(a = -16,\) \(b = 40,\) \(c = 100.\)
• Substituting these into the formula provides two solutions for \(t,\) corresponding to the times when the ball reaches exactly 100 feet.

It’s crucial to solve this correctly, as it indicates the precise points in time when the ball crosses the height threshold. By applying this formula, students build confidence in dealing with similar quadratic problems both in mathematics and in real-world scenarios involving motion.