91Ó°ÊÓ

A quadratic equation is any equation that can be written in the form of \( ax^2 + bx + c = 0 \). Here, \( a \) is the coefficient of the squared term, \( b \) is the coefficient of the linear term, and \( c \) is the constant term.
In our exercise, the height equation is given by \( s(t) = -16t^2 + 32t \), which is a quadratic equation in the form of \( at^2 + bt + c \). Notice that:

• \( a = -16 \)
• \( b = 32 \)
• \( c = 0 \)

Quadratic equations graph as parabolas. In our case, since \(a < 0\), the parabola opens downward, indicating that it has a maximum point. This is relevant because finding the maximum height of the projectile involves identifying the highest point on this parabola.

Understanding the vertex formula helps in finding the maximum or minimum value of a quadratic function. For a quadratic equation \( ax^2 + bx + c \), the vertex of the parabola occurs at \( t = -\frac{b}{2a} \).
Let's break down how we use this formula in our problem: Here, \( s(t) = -16t^2 + 32t \). We have:

• \( a = -16 \)
• \( b = 32 \)

Substituting these values into the vertex formula:
\( t = -\frac{b}{2a} \)
\( t = -\frac{32}{2(-16)} \)
\( t = 1 \)
This tells us that the object reaches its maximum height at 1 second. To find the maximum height, we substitute \( t = 1 \) back into the original height equation \( s(t) \).

Kinematics is a branch of physics that deals with the motion of objects. The height equation given in this problem is derived from kinematic equations, specifically those that describe projectile motion.
The general form of these equations incorporates the initial velocity, the effect of gravity, and time to calculate an object's position at any given moment. For an object thrown vertically upwards, the height equation is:
\( s(t) = -\frac{1}{2}gt^2 + v_0 t + s_0 \)
where:

• \( g \) is the acceleration due to gravity (32 ft/s\textsuperscript{2} in our case)
• \( v_0 \) is the initial velocity (32 ft/s here)
• \( s_0 \) is the initial position (0 ft, indicating it starts from the ground)

Plugging these values in, we get:
\( s(t) = -\frac{1}{2}(32)t^2 + 32t + 0 \)
\( s(t) = -16t^2 + 32t \). This equation tells us how the height of the object changes over time. By understanding the equations of kinematics and quadratics, we can accurately predict the behavior of projectiles.