Question

The height of a ball that is thrown straight up with a certain force is a function of the time ( \(t\) ) from which it is released given by \(f(t)=-0.5 g t^{2}+40 t\) (where \(g\) is a constant determined by gravity). a. How does the value of \(t\) at which the height of the ball is at a maximum depend on the parameter \(g\) ? b. Use your answer to part (a) to describe how maximum height changes as the parameter \(g\) changes. c. Use the envelope theorem to answer part (b) directly. d. On the Earth \(g=32,\) but this value varies somewhat around the globe. If two locations had gravitational constants that differed by \(0.1,\) what would be the difference in the maximum height of a ball tossed in the two places?

Short answer

Based on the given information, the maximum height of the ball on the moon is found to be dependent on the gravitational constant, and when comparing the heights at two different gravitational constants (32 and 32.1), the difference in maximum height is approximately 0.77 feet. The maximum height decreases as the gravitational constant increases, according to the Envelope Theorem.

Step-by-step solution

Write down the expression to represent the height of the ball as a function of time and gravitational constant

The height of the ball as a function of time and the gravitational constant is given by \(f(t)=-0.5gt^2+40t\).

Find the first derivative of the height function with respect to time

To find the maximum height, find the first derivative of the function with respect to time: \(f'(t)=-gt+40\).

Determine the time 't' when the height is maximum

The function has a maximum when the first derivative is equal to zero. Solving for \(t\) yields: \(-gt+40 = 0 \Rightarrow t = \frac{40}{g}\). #a. Analyzing the dependency of maximum height on the parameter 'g'#

Substitute time t back into the height function to obtain the maximum height

To find the height when the ball is at its maximum point, substitute the time value back into the height function: \(f\left(\frac{40}{g}\right) = -0.5g\left(\frac{40}{g}\right)^2 + 40\left(\frac{40}{g}\right)\). #b. Describing the change in the maximum height as the parameter \(g\) changes#

Simplify the maximum height expression found in the previous step

We can simplify the maximum height expression as follows: \(f\left(\frac{40}{g}\right) = -0.5g\left(\frac{1600}{g^2}\right)+\frac{1600}{\cancel{g}}g \Rightarrow f\left(\frac{40}{g}\right) = -\frac{800}{g}+1600\).

Describe how maximum height changes with the parameter g

From the simplified expression of maximum height, we observe that the maximum height decreases as the gravitational constant (\(g\)) increases. This concludes part b of our response. #c. Applying the Envelope Theorem directly#

Find the first derivative of the height function with respect to \(g\)

According to the Envelope Theorem, we need to take the derivative of the maximum height function with respect to \(g\): \(\frac{df}{dg} = -\frac{800}{g^2}\).

Describe how maximum height and the parameter \(g\) are related from the Envelope theorem

From the first derivative \(\frac{df}{dg} = -\frac{800}{g^2}\), it is clear that the maximum height decreases as the parameter \(g\) increases. This is in line with the observation made in part (b) and validates our results. #d. Calculating the difference in maximum height for two different gravitational constants #

Use the maximum height expression to find the maximum heights for both the gravitational constants

Let's consider the gravitational constants as \(g_1 = 32\) and \(g_2 = 32.1\). Maximum height expressions for both gravitational constants are: \(f\left(\frac{40}{g_1}\right) = -\frac{800}{32}+1600\) and \(f\left(\frac{40}{g_2}\right) = -\frac{800}{32.1}+1600\).

Calculate the difference between the maximum heights

Now, compute the differences between the maximum heights at both locations: \(|-\frac{800}{32} + 1600 - \left(-\frac{800}{32.1} + 1600\right)| = |\frac{800}{32} - \frac{800}{32.1}|\). After computing this expression, we get the difference in maximum height to be approximately 0.77 feet.

Key concepts

Physics of Motion

Understanding how objects move under the influence of forces, especially gravity, is a fundamental aspect of classical mechanics, a branch of physics. When a ball is thrown upwards, its motion can be described as a combination of an initial upward velocity and the constant acceleration due to gravity pulling it towards Earth.

The key to analyzing this motion lies in kinematic equations, which relate the displacement (height in this case), velocity, and acceleration of an object over time. For instance, the equation for height as a function of time used in the exercise, \(f(t)=-0.5 g t^{2}+40 t\), encapsulates these physical principles where \(g\) represents the acceleration due to gravity, and \(t\) is the time elapsed.

As the ball rises, it slows down until its velocity reaches zero at the highest point, known as the apex or the maximum height. This is the point where the ball will start its descent due to gravity's pull. An important point to remember is that the maximum height can be directly influenced by the initial force (velocity) applied and the gravity at the location where the object is thrown.

Derivative Application

Derivatives play a crucial role in physics, especially when determining the velocity and acceleration of objects in motion. When we take the derivative of the position function with respect to time, we find the velocity function. Taking the derivative of the velocity function then gives us the acceleration function. In the exercise provided, the first derivative of the height function with respect to time \(f'(t)=-gt+40\) gives us the rate of change of height, which is the velocity at time \(t\).

To find when the ball reaches its maximum height, we look for the moment when the velocity is zero — meaning the ball has momentarily stopped moving up and is about to start descending. Therefore, setting the derivative equal to zero, \(-gt+40 = 0\), allows us to find the time \(t\) at which the ball is at its peak. Substituting this time back into the original height function gives us the maximum height the ball achieves. As this solution process shows, derivatives are a powerful tool in understanding motion and predicting future states of dynamic systems.

Envelope Theorem

The Envelope Theorem is an elegant concept from economics, which finds its application in physics within problems that include parameters influencing a maximum or minimum value of a function.

In this context, the theorem helps us understand how small changes in a parameter affect the maximum height of the ball without having to calculate the maximum height from scratch for every change in that parameter. By taking the partial derivative of the maximum height with respect to the gravitational constant \(g\), denoted as \(\frac{df}{dg} = -\frac{800}{g^2}\), we can observe that the maximum height decreases as \(g\) increases.

The beauty of the Envelope Theorem is that it can provide a direct understanding of how the change in one factor influences the outcome, which in this exercise is the maximum height of a ball. This is particularly powerful because it negates the need for multiple complete calculations when only the parameter \(g\) changes.