Problem 17 - A Graphical Approach to College Algebra

Question

Solve each problem. Hitting a Baseball A baseball is hit so that its height in feet after \(t\) seconds is $$ s(t)=-16 t^{2}+44 t+4 $$ (a) How high is the baseball after 1 second? (b) Find the maximum height of the baseball.

Step-by-Step Solution

Answer

After 1 second, the baseball is 32 feet high, and its maximum height is about 32.25 feet.

Step 1: Determine the height after 1 second

To find out how high the baseball is after 1 second, substitute 1 for \( t \) in the given equation. The equation is \( s(t) = -16t^2 + 44t + 4 \). When \( t = 1 \), the equation becomes \( s(1) = -16(1)^2 + 44(1) + 4 \). Simplify to find \( s(1) = -16 + 44 + 4 = 32 \). Thus, the baseball is 32 feet high after 1 second.

Step 2: Find the vertex of the parabola

The function \( s(t) = -16t^2 + 44t + 4 \) is a quadratic equation, which represents a parabola. The maximum height of the baseball corresponds to the vertex of this parabola. The \( t \)-coordinate of the vertex can be found using the formula \( t = -\frac{b}{2a} \), where \( a = -16 \) and \( b = 44 \). Substitute these values into the formula: \( t = -\frac{44}{2(-16)} = \frac{44}{32} = 1.375 \).

Step 3: Calculate the maximum height

Now, substitute \( t = 1.375 \) into the original equation \( s(t) = -16t^2 + 44t + 4 \) to find the maximum height. Calculate \( s(1.375) = -16(1.375)^2 + 44(1.375) + 4 \). Simplifying, we find \( s(1.375) = -16(1.890625) + 60.5 + 4 \approx 32.25 \). Therefore, the maximum height of the baseball is approximately 32.25 feet.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a crucial concept when working with quadratic equations. A quadratic equation, like the one seen in the baseball problem, generally has the form \( ax^2 + bx + c \). This equation represents a parabola, which can either open upwards or downwards depending on the value of \( a \). If \( a \) is positive, the parabola opens upwards, and if \( a \) is negative, it opens downwards.
For the equation \( s(t) = -16t^2 + 44t + 4 \), the parabola opens downwards because \( a = -16 \). This means the vertex represents the maximum point of the parabola, and hence, the maximum height of the baseball in our context. To find the vertex, we use the formula \( t = -\frac{b}{2a} \). Here, \( b = 44 \) and \( a = -16 \).
Substituting these values gives us \( t = \frac{44}{32} = 1.375 \). This means the vertex, or the time at which the baseball reaches its maximum height, is at \( t = 1.375 \) seconds.

Maximum Height

The maximum height of a projectile like a baseball is an important element of understanding its trajectory. In our baseball scenario, the height corresponds to the vertex of the parabola described by the equation \( s(t) = -16t^2 + 44t + 4 \).
Once the time \( t \) at which the maximum height occurs is determined (as \( t = 1.375 \) from the vertex formula), we can substitute this back into the quadratic equation to find the maximum height. The calculation goes like this:

• First, plug \( t = 1.375 \) into the equation: \( s(1.375) = -16(1.375)^2 + 44(1.375) + 4 \).
• Calculate \( (1.375)^2 = 1.890625 \) and then \( -16 \times 1.890625 = -30.25 \).
• Add \( 44 \times 1.375 = 60.5 \) and finally \( +4 \).

Adding these values together yields \( s(1.375) \approx 32.25 \) feet. Therefore, the maximum height the baseball reaches is approximately 32.25 feet.

Quadratic Formula

Quadratic equations are often solved using the quadratic formula, an indispensable tool in algebra. A quadratic equation in standard form is given by \( ax^2 + bx + c = 0 \). The quadratic formula then allows us to find the values of \( x \) that satisfy this equation and is expressed as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]However, in the baseball problem, we're not solving for zeros or roots of the equation, but rather for the vertex and maximum height, which do not primarily use the quadratic formula.
Instead, we utilize the vertex formula \( t = -\frac{b}{2a} \). The quadratic formula is typically employed when you need to find when a projectile returns to its original height or makes contact with the ground, which would be the roots or solutions of the equation \( s(t) = 0 \).
Understanding when and how to apply these various formulas is critical in solving problems involving quadratic equations effectively.