Problem 37 - Thomas Calculus

Question

Volleyball A volleyball is hit when it is 4 ft above the ground and 12 ft from a 6 -ft-high net. It leaves the point of impact with an initial velocity of 35 \(\mathrm{ft} / \mathrm{sec}\) at an angle of \(27^{\circ}\) and slips by the opposing team untouched. a. Find a vector equation for the path of the volleyball. b. How high does the volleyball go, and when does it reach maximum height? c. Find its range and flight time. d. When is the volleyball 7 ft above the ground? How far e. Suppose that the net is raised to 8 ft. Does this change things? Explain.

Step-by-Step Solution

Answer

The volleyball follows a parabolic path, reaching a max height at specific \( t \), and clearing the net if it reaches necessary height before passing over.

Step 1: Understand the Initial Conditions

The volleyball is hit 4 ft above the ground and 12 ft from the net which is 6 ft tall. Its initial velocity is 35 ft/sec at an angle of 27°, which can be broken into horizontal and vertical components using trigonometry.

Step 2: Equations for Horizontal and Vertical Motion

For horizontal motion, use: \[ x(t) = v_x \, t + x_0 \] where \(v_x = 35 \cos(27°)\) and \(x_0 = 12\).For vertical motion, use: \[ y(t) = -16t^2 + v_y \, t + y_0 \] where \(v_y = 35 \sin(27°)\) and \(y_0 = 4\).

Step 3: Find the Vector Equation for the Path

The vector equation can be expressed as: \[ \vec{r}(t) = \begin{pmatrix} x(t) \ y(t) \end{pmatrix} = \begin{pmatrix} 35 \cos(27°) \, t + 12 \ -16t^2 + 35 \sin(27°) \, t + 4 \end{pmatrix} \]

Step 4: Determine Maximum Height and Time

The maximum height is found using the vertex of the parabola in the vertical motion equation. Set the derivative to zero: \[ \frac{dy}{dt} = -32t + 35 \sin(27°) = 0 \]Solve for \(t\) to get the time to maximum height, then substitute this into \(y(t)\) to find the maximum height.

Step 5: Calculate Range and Flight Time

The range can be found by solving \( y(t) = 0 \) for the time \( t \) when the ball returns to the original vertical height of 4 ft. Substitute this \( t \) back into the \( x(t) \) equation to find the range.

Step 6: Find Height of 7 ft and Corresponding Horizontal Distance

Solve \( y(t) = 7 \) for \( t \), using the quadratic formula as needed. Substitute these times into \( x(t) \) to find the horizontal distance when the ball is 7 ft above the ground.

Step 7: Evaluate the Impact of a 8 ft Net

Determine if and when the ball surpasses 8 ft by checking the height and time when the net would be in the ball's path. Analyze the equation of motion to see if and when \( y(t) = 8 \) occurs before \( x(t) \) equals the net distance.

Key Concepts

Vector Equations

In the realm of projectile motion, vector equations are powerful tools used to describe the path of an object in motion. A vector equation essentially combines information about both the horizontal and vertical components of motion into a cohesive mathematical statement. For a volleyball hit into the air, this equation characterizes how its position changes over time.

Using the initial velocity and angle of launch, we can break the volleyball's motion into horizontal and vertical components:

• Horizontal component: \(v_x = v_0 \cos(\theta)\)
• Vertical component: \(v_y = v_0 \sin(\theta)\)

In our volleyball example:

• Initial velocity \(v_0 = 35\,\text{ft/sec}\)
• Launch angle \(\theta = 27^\circ\)
• Starting horizontal position \(x_0\) = 12 ft
• Starting vertical position \(y_0\) = 4 ft

This leads to the vector equation \( \vec{r}(t) = \begin{pmatrix} x(t) \ y(t) \end{pmatrix} = \begin{pmatrix} 35 \cos(27^\circ) \, t + 12 \ -16t^2 + 35 \sin(27^\circ) \, t + 4 \end{pmatrix}\). This equation gives a full representation of the motion path.

Trigonometry

Trigonometry plays a crucial role in determining the components of projectile motion. When a volleyball is launched at a particular angle, its velocity can be split into two distinct components using trigonometric functions.

For instance, the angle of launch (\(27^\circ\) in this case) helps us find:

• The horizontal velocity component: \(v_x = v_0 \cos(\theta)\). This component dictates how quickly the volleyball moves across the court.
• The vertical velocity component: \(v_y = v_0 \sin(\theta)\). This determines how the volleyball rises and falls due to gravity.

These trigonometric expressions are derived from the standard usage of sine and cosine functions for acute angles, where \(\cos(\theta)\) complements \(\sin(\theta)\) between the opposite and adjacent sides of a right triangle.This breakdown is vital for accurately modeling and predicting the path the volleyball will take.

Parabolic Motion

The projectile motion of a volleyball can be described as parabolic. This means it follows a symmetrical, curved path, typical of objects propelled into the air under the influence of gravity.

In our example, the equation for vertical motion is \( y(t) = -16t^2 + v_y \, t + y_0 \), which resembles the equation for a parabola, \( f(t) = at^2 + bt + c \). Here, the \(-16t^2\) term represents the constant acceleration due to gravity, pulling the ball down at 32 ft/sec² (halved since it's \' ft in feet compared to \(-32 \text{ft/s}^2\) directly in time units squared).

The parabolic shape arises from the balance of:

• Initial upward velocity due to the angle of launch.
• The constant downward pull of gravity.

This classic behavior of motion is described by the path formula in the vertical direction and is crucial in predicting where the ball lands.

Maximum Height

In projectile motion, the maximum height represents the highest vertical point reached by an object. For the volleyball, we calculate this using the time when the vertical speed becomes zero.

The derivative of the vertical position function with respect to time \( \frac{dy}{dt} = -32t + 35 \sin(27^\circ) \) is key in identifying the maximum height. We set this derivative to zero because the peak occurs when the vertical velocity is zero:

\(-32t + 35 \sin(27^\circ) = 0\)

• Solve for \(t\) to find when the volleyball reaches its maximum height.
• Substitute this \(t\) back into \(y(t) = -16t^2 + 35 \sin(27^\circ) \, t + 4\) to find the actual height.

This gives a precise understanding of how long the volleyball ascends before gravity pushes it down again.

Rational Roots

Finding rational roots can be an essential part of understanding when a certain height is reached during projectile motion. When you need to calculate precise instances at which the volleyball reaches a given height, like 7 ft, you might need to solve a quadratic equation.

This equation is formed by setting the vertical motion equation to the desired height, hence solving \(y(t) = 7\). The quadratic formula helps navigate through situations where simple roots aren't apparent directly.

• A quadratic equation \(at^2 + bt + c = 0\) has solutions given by the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Each solution corresponds to a specific time when the volleyball reaches the targeted height.

Recognizing when these rational roots appear allows for the precise prediction of motion paths, particularly if needing to check if the ball clears a certain height such as the 8-ft net.