Problem 33 - 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 8 ft. Does this change things? Explain.

Step-by-Step Solution

Answer

The volleyball's path is described by given vector equation, it peaks at 7.95 ft at 0.497 sec, has a range of 40.37 ft in 1.29 sec, and is 7 ft high at 0.31 and 0.61 sec. Raising the net to 8 ft affects the shot.

Step 1: Understand the Problem

The volleyball is hit when it is 4 ft above the ground and travels with an initial velocity of 35 ft/sec at an angle of 27°. We need to find a vector equation for its path, its maximum height and when it reaches it, its range and flight time, when it is 7 ft above ground, and discuss the impact of raising the net height.

Step 2: Find Initial Position and Velocity Components

The initial position of the volleyball is (0, 4) since it is 4 ft above the ground. The initial velocity components in the horizontal \( v_{0x} \) and vertical \( v_{0y} \) directions are calculated using:\[ v_{0x} = 35 \cos(27^{\circ}) \approx 31.29 \, \text{ft/sec}, \ v_{0y} = 35 \sin(27^{\circ}) \approx 15.90 \, \text{ft/sec}.\]

Step 3: Write the Vector Equation

The vector equation for the motion of the volleyball using the initial position and velocity components is:\[ \begin{align*} x(t) &= v_{0x} t = 31.29 t, \ y(t) &= 4 + v_{0y} t - \frac{1}{2}gt^2 \ &\approx 4 + 15.90t - 16t^2, \end{align*}\]where \( g = 32 \, \text{ft/sec}^2 \) is the acceleration due to gravity.

Step 4: Determine Maximum Height and Time at Maximum Height

The maximum height is found by setting the vertical velocity to zero: \[ v_y = v_{0y} - gt = 0. \ t = \frac{v_{0y}}{g} = \frac{15.90}{32} \approx 0.497 \text{ sec}.\]Substitute \( t \) in \( y(t) \) to find the height:\[ y(0.497) = 4 + 15.90 \times 0.497 - 16 \times (0.497)^2 \approx 7.95 \text{ ft}.\]

Step 5: Calculate Range and Flight Time

The flight ends when the volleyball hits the ground \((y = 0)\). Solving for \( t \) in\[ 0 = 4 + 15.9t - 16t^2 \ t \approx 1.29 \, \text{sec}.\]The range \( x \) at this time is:\[ x(1.29) = 31.29 \times 1.29 \approx 40.37 \text{ ft}.\]

Step 6: Find When Volleyball is 7 ft High

Set \( y(t) = 7 \) and solve for \( t \):\[ 7 = 4 + 15.9t - 16t^2 \ 16t^2 - 15.9t + 3 = 0.\]Solving this quadratic, \( t \approx 0.31 \text{ sec} \) or \( t \approx 0.61 \text{ sec} \).

Step 7: Effect of Raising the Net

If the net height is increased to 6 ft or more, compare with the 7.95 ft peak height. Since the new net is 8 ft and exceeds 7.95 ft, the ball hits the net, affecting its trajectory and making a different scenario than the one initially described.

Key Concepts

Vector Equation

In the context of projectile motion, a vector equation describes the volleyball's path. It essentially captures the relationship between time and the particle's position in space. For the volleyball in this exercise, the vector equation is derived by considering its initial conditions and the forces acting on it, primarily gravity.
The initial position is given as 4 feet above ground, while the launch angle and initial velocity are known. We use these to find the components of velocity. Employing trigonometric functions, we separate the velocity into horizontal and vertical components:

• The horizontal velocity component: \( v_{0x} = v_0 \cos(\theta) = 35 \cos(27^\circ) \approx 31.29 \, \text{ft/sec} \)
• The vertical velocity component: \( v_{0y} = v_0 \sin(\theta) = 35 \sin(27^\circ) \approx 15.90 \, \text{ft/sec} \)

The vector equation for motion becomes:

• Horizontal position as a function of time: \( x(t) = v_{0x} t \)
• Vertical position as a function of time: \( y(t) = 4 + v_{0y} t - \frac{1}{2}gt^2 \)

Here \( g \) represents the gravitational acceleration, approximated as \( 32 \, \text{ft/sec}^2 \). This equation provides a full representation of the volleyball's path as influenced by its initial conditions and the constant force of gravity.

Maximum Height

The maximum height of a projectile is the highest vertical position it reaches during its flight. For the volleyball, this is found by examining when its vertical velocity is zero. To express this mathematically, set the derivative of the vertical position equation (which gives vertical velocity) to zero:\[ v_y = v_{0y} - gt = 0 \]We solve for \( t \) to determine when this occurs:\[ t = \frac{v_{0y}}{g} = \frac{15.90}{32} \approx 0.497 \text{ sec} \]After finding the time, plug it into the vertical motion equation \( y(t) \) to find the maximum height:\[ y(0.497) = 4 + 15.90 \times 0.497 - 16 \times (0.497)^2 \approx 7.95 \text{ ft} \]Therefore, the volleyball reaches its peak height, approximately 7.95 feet above the ground, around 0.497 seconds after being hit. This is an essential part of assessing whether the ball clears particular obstacles, like a net in volleyball.

Flight Time

Flight time is the duration the volleyball stays in the air. It is determined by identifying when the volleyball once again reaches the ground level (\( y=0 \)). The vertical motion equation \( y(t) = 4 + v_{0y} t - \frac{1}{2}gt^2 \) is used to solve for the time \( t \) when the volleyball hits the ground:Set \( y(t) = 0 \): \[ 0 = 4 + 15.9t - 16t^2 \]This is a quadratic equation in terms of \( t \). Solving it using algebraic methods gives:\[ t \approx 1.29 \, \text{sec} \]This indicates that the volleyball's total time in the air is approximately 1.29 seconds before it touches down. This time frame encompasses the entire trajectory starting from the ball being hit to coming back to the ground level. Understanding flight time helps in analyzing the complete motion and the effective range the volleyball can cover.

Quadratic Equation

Quadratic equations often appear in projectile motion problems because the vertical position equation is, indeed, a quadratic equation. In this scenario, you can see this pattern in the vertical motion expression:\[ y(t) = 4 + 15.9t - 16t^2 \]Recognizing the equation's form is crucial for solving projectile questions such as determining when the ball reaches specific heights or lands. For example, finding when the volleyball is 7 feet above ground level involves solving:\[ 7 = 4 + 15.9t - 16t^2 \]Or equivalently:\[ 16t^2 - 15.9t + 3 = 0 \]Which is another quadratic equation. Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 16 \), \( b = -15.9 \), \( c = 3 \), we solve for \( t \). The solutions are \( t \approx 0.31 \text{ sec} \) or \( t \approx 0.61 \text{ sec} \), indicating times when the volleyball reaches this particular height.Quadratic equations provide a systematic way to determine various aspects of projectile motion. They help describe the parabolic trajectory inherent to these kinds of motions.