Problem 59 - Algebra and Trigonometry

Question

Height of a Ball 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: Understand the problem

The problem asks us to find the maximum height attained by the ball thrown directly upward with a given velocity. The height as a function of time is provided as \[ y(t) = 40t - 16t^2. \]

Step 2: Identify the type of function

The given function is\[ y(t) = 40t - 16t^2 \]which is a quadratic equation in the standard form \[ at^2 + bt + c, \] where \( a = -16, b = 40, \) and \( c = 0. \) This represents a downward-opening parabola since \( a \) is negative.

Step 3: Find the vertex of the parabola

The vertex of a parabola given by the equation \( at^2 + bt + c \) occurs at \( t = -\frac{b}{2a}. \) Substituting \( a = -16 \) and \( b = 40 \) into this formula,\[ t = -\frac{40}{2(-16)} = \frac{40}{32} = \frac{5}{4}. \]

Step 4: Calculate the maximum height

Substitute \( t = \frac{5}{4} \) back into the height equation:\[ y\left(\frac{5}{4}\right) = 40\left(\frac{5}{4}\right) - 16\left(\frac{5}{4}\right)^2. \]Calculate each term separately:\[ 40\left(\frac{5}{4}\right) = 50 \]and \[ 16\left(\frac{5}{4}\right)^2 = 16\left(\frac{25}{16}\right) = 25. \]So, \[ y\left(\frac{5}{4}\right) = 50 - 25 = 25. \]

Step 5: Conclusion

The maximum height attained by the ball is \( 25 \) feet.

Key Concepts

Parabolas

A parabola is a U-shaped curve that can either open upwards or downwards. In our exercise, we're dealing with an equation that forms a downward-opening parabola.
Parabolas are significant in math because they represent quadratic relationships. The general form of a quadratic equation is \( y = ax^2 + bx + c \). Here, \( a \), \( b \), and \( c \) are constants, each controlling different aspects of the parabola's shape and position.

• When \( a > 0 \), the parabola opens upwards like a cup.
• When \( a < 0 \), it opens downwards like an upside-down cup.

This particular feature determines if the parabola has a maximum or minimum point. In our example, since \( a = -16 \), the parabola opens downwards, suggesting it has a maximum point.

Vertex of a Parabola

The vertex of a parabola is its turning point. It's where the curve changes direction.
In the context of a quadratic equation, the vertex is significant because it represents the maximum or minimum value, depending on whether the parabola opens upwards or downwards.
For a quadratic equation \( y = ax^2 + bx + c \), the formula to find the vertex \( t \)-coordinate is \( t = -\frac{b}{2a} \).
In our exercise, using the equation \( y = 40t - 16t^2 \), we determined:

• \( a = -16 \)
• \( b = 40 \)

By substituting these into our vertex formula, \( t = \frac{5}{4} \). This means the ball reaches the highest point at \( t = \frac{5}{4} \) seconds.

Maximum Height Calculation

Calculating the maximum height involves finding the vertex's corresponding \( y \)-value, because in a downward-opening parabola, this point is at the peak.
From our previous calculation, we found the time \( t = \frac{5}{4} \) seconds where the maximum height occurs. To find this maximum height, we substitute \( t = \frac{5}{4} \) back into the original height equation:
\[ y\left(\frac{5}{4}\right) = 40\left(\frac{5}{4}\right) - 16\left(\frac{5}{4}\right)^2 \]
We calculate each term separately:

• The first part: \( 40\left(\frac{5}{4}\right) = 50 \)
• The second part: \( 16\left(\frac{5}{4}\right)^2 = 16\left(\frac{25}{16}\right) = 25 \)

Substituting these calculations back, we find:
\[ y\left(\frac{5}{4}\right) = 50 - 25 = 25 \]
Thus, the maximum height of the ball is 25 feet.