91Ó°ÊÓ

When working with problems involving vertical motion, the idea of a derivative becomes crucial, especially in the realms of physics and calculus. The derivative, in simple terms, represents the rate of change. In our exercise, the height of the object with respect to time is given by the quadratic function: \[ s(t) = -16t^2 + 96t + 112. \] To find out how fast the object is moving at any given moment, we derive this function with respect to time \( t \). This process of differentiation gives us the velocity function: \[ v(t) = \frac{ds}{dt} = -32t + 96. \] This new function tells us the velocity of the object at any point in time. Differentiating a quadratic equation like this one is straightforward. You just multiply the coefficient by the power of the term and then decrease its power by one. Consequently, the derivative of a position function provides us insight into motion by revealing the velocity.

The maximum height of an object in vertical motion is a key concept, particularly in physics. For our given quadratic equation \( s(t) = -16t^2 + 96t + 112 \), the highest point, or the maximum height, occurs at the vertex of the parabola. In mathematics, the vertex of a parabola described by the function \( at^2 + bt + c \) is found using the formula: \[ t = -\frac{b}{2a}. \] For the function in our exercise, where \( a = -16 \) and \( b = 96 \), we find that the time at which the maximum height occurs is: \[ t = -\frac{96}{2(-16)} = 3 \text{ seconds}. \] To find the actual maximum height, substitute this time back into the original height function: \[ s(3) = -16(3)^2 + 96(3) + 112 = 256 \text{ feet}. \] This simple substitution reveals that the object reaches its maximum height of 256 feet after 3 seconds.

In the realm of vertical motion, the quadratic formula is often used to find the time when the object in motion reaches a specific height, such as when it hits the ground. The equation \[ s = -16t^2 + 96t + 112 \] represents the height. To determine when the object reaches the ground, set \( s \) to zero and solve for \( t \): \[ 0 = -16t^2 + 96t + 112. \] The quadratic formula, \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]helps us find the solution for \( t \). In our problem, the coefficients are: \( a = -16, \) \( b = 96,\) and \( c = 112. \) Plugging these into the formula gives: \[ t = \frac{-96 \pm \sqrt{96^2 + 4 \times 16 \times 112}}{-32}. \]After simplification, this results in approximately \( t \approx 7 \text{ seconds}. \) This solution indicates that the object reaches the ground 7 seconds after being released.

The concept of the velocity function is central to understanding motion. In our context, the velocity function tells us how the speed of the object changes over time. From the height equation, we found the velocity function via differentiation:\[ v(t) = -32t + 96. \]This equation allows you to plug in any value of \( t \) and immediately know the velocity or speed of the object at that particular moment. Let's explore two specific instances:

• At \( t = 0 \), the velocity is given by \( v(0) = -32(0) + 96 = 96 \text{ feet per second}. \) This indicates the initial speed upward.
• When the object reaches the ground at \( t = 7 \text{ seconds} \), the velocity becomes \( v(7) = -32(7) + 96 = -128 \text{ feet per second}. \) This implies a downward speed as it hits the ground.

Understanding the velocity function provides keen insight into how and when the speed changes, highlighting key moments in the object's trajectory.