Problem 70 - Calculus Single Variable

Question

At Wrigley field in Chicago, Cubs fans throw the ball back onto the field when a visiting team hits a home run. Suppose that the height \(H\) above field level of a ball thrown back by a Cubs fan is given in feet by $$H(t)=18+13.8 t-16 t^{2}$$ when \(t\) is measured in seconds. a. How high above field level is the fan sitting? b. What is the rate at which the ball rises as a function of time? c. At what time does the ball reach its maximum height? What is this maximum height? d. What is the average rate of change of \(H\) during the ball's upward trajectory? e. How long is the ball in the air? f. What is the rate of change of \(H\) at the moment the ball hits the ground? g. What is the average rate of change of \(H\) over the entire trajectory of the ball?

Step-by-Step Solution

Answer

a. 18 feet; b. \(H'(t) = 13.8 - 32t\); c. 0.431 sec, 21.971 feet; d. 9.205 ft/s; e. 1.275 sec; f. -27 ft/s; g. -14.118 ft/s.

Step 1: Determine the Fan's Sitting Height

The initial height above field level can be found by evaluating the height function at time \( t = 0 \). This gives us:\[ H(0) = 18 + 13.8(0) - 16(0)^2 = 18 \text{ feet}. \] So, the fan is sitting 18 feet above field level.

Step 2: Calculate the Rate of Change of Height Over Time

The rate of change of height, or velocity, as a function of time can be found by taking the derivative of \( H(t) \):\[ H'(t) = \frac{d}{dt}(18 + 13.8t - 16t^2) = 13.8 - 32t. \] Thus, the velocity function is \( H'(t) = 13.8 - 32t \).

Step 3: Find the Time and Value of Maximum Height

To find the time of maximum height, set \( H'(t) = 0 \):\[ 13.8 - 32t = 0 \quad \Rightarrow \quad t = \frac{13.8}{32} = 0.43125 \text{ seconds}. \] Substituting \( t = 0.43125 \) back into \( H(t) \) gives:\[ H(0.43125) = 18 + 13.8(0.43125) - 16(0.43125)^2 = 21.971 \text{ feet}. \] Thus, the maximum height is approximately 21.971 feet.

Step 4: Compute the Average Rate of Change Upwards

The ball's upward trajectory starts at \( t = 0 \) and reaches a maximum at \( t = 0.43125 \). The average rate of change is:\[ \frac{H(0.43125) - H(0)}{0.43125 - 0} = \frac{21.971 - 18}{0.43125} \approx 9.205 \text{ ft/s}. \] This is the average rate of change during the upward motion.

Step 5: Determine Total Air Time

The ball is in the air until it hits the ground. Set \( H(t) = 0 \) and solve for \( t \):\[ 0 = 18 + 13.8t - 16t^2. \] Solving the quadratic equation using the quadratic formula gives two solutions; discard the negative one: \( t = 1.275 \text{ seconds}. \) Thus, the ball is in the air for about 1.275 seconds.

Step 6: Rate of Change of Height When Ball Hits Ground

Using the velocity function \( H'(t) = 13.8 - 32t \), substitute \( t = 1.275 \):\[ H'(1.275) = 13.8 - 32(1.275) = -27 \text{ ft/s}. \] The velocity is \(-27\) ft/s when the ball hits the ground.

Step 7: Average Rate of Change Over Entire Trajectory

The average rate of change from launch to landing is:\[ \frac{H(1.275) - H(0)}{1.275 - 0} = \frac{0 - 18}{1.275} \approx -14.118 \text{ ft/s}. \] This is the average rate of change over the ball’s entire flight.

Key Concepts

Quadratic Functions

The concept of quadratic functions is central to understanding projectile motion. A quadratic function can be expressed in the form of \(f(t) = at^2+bt+c\). In projectile motion, this formula is used to describe the path of objects moving through the air under the force of gravity and is depicted as a parabola.
In the context of our exercise, the function given is \(H(t) = 18 + 13.8t - 16t^2\). Here:

• The coefficient \(a = -16\) represents the effect of gravity on the projectile's motion. Notice that it is negative, consistent with the direction of gravitational pull downwards.
• The coefficient \(b = 13.8\) indicates the initial velocity with which the ball is thrown upwards by the fan.
• The constant \(c = 18\) represents the fan's sitting height above field level when the ball was released.

This function provides a predictive model of the ball's height at any given time \(t\). By manipulating quadratic functions, we can find key characteristics of the ball's trajectory, such as its maximum height and how long it is in the air.

Derivatives

A derivative in mathematics provides a way to determine the rate at which a quantity changes. In our context, the derivative of the height function \(H(t)\) is crucial since it gives us the velocity of the ball at any time.
The derivative \(H'(t)\) is found by differentiating the original quadratic function \(H(t) = 18 + 13.8t - 16t^2\):
\[ H'(t) = \frac{d}{dt}(18 + 13.8t - 16t^2) = 13.8 - 32t. \]
This new function describes how fast the ball is moving at any given moment. Specifically:

• If \(H'(t) > 0\), the ball is rising.
• If \(H'(t) < 0\), the ball is falling.
• If \(H'(t) = 0\), the ball is at its peak height.

In our exercise, the derivative is leveraged to find when the ball reaches its maximum height and the rate of change as the trajectory progresses.

Maximum Height

The concept of maximum height is essential because it represents the pinnacle of the ball's arc in projectile motion. Achieving maximum height occurs when the upward velocity of an object becomes zero.
In mathematical terms, this is when the derivative of the height function \(H'(t)\) is zero:
\[ 13.8 - 32t = 0. \]
Solving for \(t\) gives \( t = \frac{13.8}{32} = 0.43125 \text{ seconds}\).
To find the maximum height, substitute this \(t\)-value back into the original function \(H(t)\):
\[ H(0.43125) = 18 + 13.8(0.43125) - 16(0.43125)^2 \approx 21.971 \text{ feet}. \]
This means the ball, at its peak, is approximately 21.971 feet above the field. Understanding how to find the maximum height is crucial as it provides insights into the behavior of any projectile.

Average Rate of Change

The average rate of change offers a broader view of how a function behaves over a certain interval. For the ball's trajectory, it tells us the mean rate at which the ball's height changes.
For a given time interval \((t_1, t_2)\), the average rate of change of the height function \(H(t)\) is calculated by:
\[ \frac{H(t_2) - H(t_1)}{t_2 - t_1}. \]
In the exercise, to determine the ball's average rate of change during its upward motion from \(t = 0\) to \(t = 0.43125\):
\[ \frac{H(0.43125) - H(0)}{0.43125 - 0} = \frac{21.971 - 18}{0.43125} \approx 9.205 \text{ ft/s}. \]
This metric shows the ball's average rising speed from launch to peak. The method remains the same for other periods, such as the entire trajectory of the ball, providing insights into overall performance.