Problem 63 - Precalculus Mathematics for Calculus

Question

If a ball is thrown directly upward with a velocity of \(40 \mathrm{ft} / \mathrm{s}\), its height (in feet) after \(t\) seconds is given by \(y=40 t-16 t^{2} .\) What is the maximum height attained by the ball?

Step-by-Step Solution

Answer

The maximum height attained by the ball is 25 feet.

Step 1: Understanding the Equation

The height of the ball is described by the equation \(y = 40t - 16t^2\), where \(y\) is the height in feet, and \(t\) is the time in seconds. This is a quadratic equation in the form \(ax^2 + bx + c\), where \(a = -16\), \(b = 40\), and \(c = 0\).

Step 2: Identifying the Maximum Height

The maximum height of a projectile can be found using the vertex of the parabola represented by the quadratic equation. The time at which the maximum height occurs is given by \(t = -\frac{b}{2a}\).

Step 3: Calculating the Time of Maximum Height

Substitute the values of \(a\) and \(b\) into the formula: \(t = -\frac{40}{2(-16)} = \frac{40}{32} = 1.25\) seconds.

Step 4: Finding the Maximum Height

Substitute \(t = 1.25\) back into the height equation: \[y = 40(1.25) - 16(1.25)^2\]Calculate the values:\[y = 50 - 16(1.5625)\]\[y = 50 - 25 = 25\]Thus, the maximum height is 25 feet.

Key Concepts

Projectile Motion

Projectile motion deals with objects being launched into the air and follows the laws of physics. In our exercise, a ball is thrown upward, creating a specific type of motion called **projectile motion**. This motion is governed by the forces acting on the object, such as gravity, which causes the ball to eventually slow down and fall back to the ground. Important aspects to consider in projectile motion include:

• The initial velocity, which is the speed at which the object is launched. Here, it's 40 feet per second upwards.
• The influence of gravity, which consistently acts downwards, altering the object's path. In our problem, this force is represented by the \(-16t^2\) term in the equation, responsible for creating the parabolic shape of the trajectory.

Using these considerations, the motion of any projectile can be represented by a quadratic equation, which allows us to predict things like maximum height and time in the air.

Maximum Height

In projectile motion, the maximum height is the apex of the trajectory—the highest point the object reaches.To find this, we need to determine when the ball stops moving upward and starts descending. This happens exactly at the maximum height, where the velocity temporarily equals zero.From the provided formula, the maximum height can be calculated using the concept of the parabola's vertex. The time at which this occurs is calculated as:\[ t = -\frac{b}{2a} \]Given the values in the problem, where \( b = 40 \) and \( a = -16 \), this transforms to:\[ t = -\frac{40}{2 \times (-16)} = \frac{40}{32} = 1.25 \text{ seconds} \]This means that after 1.25 seconds, the ball reaches its maximum height. By substituting this time back into the initial height equation:\[ y = 40(1.25) - 16(1.25)^2 \]we find that the height \( y \) reaches 25 feet, confirming it as the maximum height.

Parabola Vertex

The vertex of a parabola holds crucial information about its shape and orientation.In the context of our problem, the vertex tells us both the time and height at which the ball reaches its maximum height.Every quadratic equation of the form \( ax^2 + bx + c \) has a vertex at:\[ \left(-\frac{b}{2a}, y(-\frac{b}{2a})\right) \]This represents the point \( (t, y) \), where the parabola changes direction. For our specific exercise, inserting the known values \( b = 40 \) and \( a = -16 \), we calculated the time \( t = 1.25 \) seconds as the moment the ball reaches its maximum height. By plugging this value back into the equation, we confirmed that the height \( y \) at this point is 25 feet. The vertex represents the pinnacle of the projectile's path, serving as a valuable insight into its trajectory and helps to confirm our calculations in projectile motion problems.