Problem 27 - 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 \(\mathrm{ft}\) above the ground? How far (ground distance) is the volleyball from where it will land? e. Suppose that the net is raised to 8 \(\mathrm{ft}\) . Does this change things? Explain.

Step-by-Step Solution

Answer

Use trigonometry to split velocity into components; solve quadratic for max height, time, and range.

Step 1: Define Initial Conditions and Vectors

We have the initial point of impact as (0, 4) because the volleyball is hit when it's 4 ft above the ground. The initial velocity is given as 35 ft/sec at an angle of 27°. We need to find the horizontal and vertical components of this velocity using trigonometry:\[v_x = 35 \cos(27^") \v_y = 35 \sin(27^")\]

Step 2: Derive the Vector Equation for the Path (Part a)

Using the components from Step 1, the vector equation describing the path of the projectile can be given by:\[\begin{align*}x(t) &= v_x t = 35 \cos(27^") t \y(t) &= -\frac{1}{2}gt^2 + v_y t + 4 = -\frac{1}{2}(32)t^2 + 35 \sin(27^") t + 4\end{align*}\]where \( g = 32 \text{ ft/sec}^2 \) is the acceleration due to gravity.

Step 3: Calculate Maximum Height and Time (Part b)

The vertex of the parabola (maximum height) occurs at \( t = \frac{v_y}{g} = \frac{35 \sin(27^")}{32} \). Use this time in the "y" component of the path equation to find the maximum height.\[h_{\text{max}} = -\frac{1}{2}(32) \left(\frac{35 \sin(27^")}{32}\right)^2 + 35 \sin(27^") \left(\frac{35 \sin(27^")}{32}\right) + 4\]

Step 4: Determine Range and Flight Time (Part c)

The flight time can be found by solving the equation \( y(t) = 0 \) for \( t \). Substitute this into \( x(t) = 35 \cos(27^") t \) to find the range, which is the horizontal distance when the ball lands.

Step 5: Find When Volleyball is 7 ft Above Ground (Part d)

Solve \( y(t) = 7 \) for \( t \) using the "y" equation. Then, use these \( t \) values in the "x" equation to find how far the volleyball travels when it is 7 ft above the ground, and determine the ground distance from landing.

Step 6: Evaluate Raising Net Height to 8 ft (Part e)

Check if the volleyball height (\(y(t)\)) when it is 12 ft away (the distance to the net \(x(t)\)) is greater than 8 ft. If the height is below 8 ft at this point, the volleyball will hit the net.

Key Concepts

Vector Equation

Projectile motion can be illustrated with a vector equation that describes the path of an object as it moves through the air. For a volleyball, as in our example, this path is determined by its initial velocity and the angle at which it is hit. The vector equation combines horizontally and vertically oriented components, which are affected by initial speed, launch angle, and gravitational pull.

• Horizontal Component: The horizontal component of the velocity is calculated using the formula: \( v_x = v \cdot \cos(\theta) \). This determines how far the volleyball will travel in the horizontal direction.
• Vertical Component: The vertical velocity is given by \( v_y = v \cdot \sin(\theta) \). This affects how high and for how long the volleyball will stay airborne.

To develop the vector equation, you formulate two equations:

• \( x(t) = v_x \cdot t \)
• \( y(t) = -\frac{1}{2}gt^2 + v_y \cdot t + y_0 \)

Here, \( g \) is the acceleration due to gravity, typically \(32 \text{ ft/s}^2\), and \( y_0 \) is the initial height from where the projectile starts.

Maximum Height

Determining the maximum height reached by a projectile, like a volleyball, involves analyzing the vertical motion. The maximum height is achieved when the vertical component of the velocity is zero, which is the peak of the projectile's arc.To find this, use the vertical component of the velocity \( v_y \). The time to reach maximum height \( t_{max} \) is:\[ t_{max} = \frac{v_y}{g} \]Using our example, once you've calculated \( t_{max} \), substitute it back into the vertical position equation \( y(t) \):\[ h_{max} = -\frac{1}{2}gt_{max}^2 + v_y \cdot t_{max} + y_0 \]By calculating this, you determine how high the volleyball climbs before descending, which is critical for understanding its performance during a game.

Flight Time

The total time the volleyball remains in the air is known as the flight time. This duration is crucial in sports like volleyball, as it influences strategies and responses of the opposing players. Flight time is determined by solving the vertical motion equation for when the projectile returns to the ground level \( y(t) = 0 \).
To find it:

• Solve the equation for \( y(t) = 0 \), yielding the time \( t \) when the volleyball hits back on the ground.

Plug this time value into the horizontal motion equation to find out the final location or range. This process highlights the interaction between time spent in the air and horizontal distance covered.

Range

The range of a projectile is the horizontal distance it covers during its entire flight until it hits the ground. Calculating the range involves both the initial speed and the angle of launch. Once you have determined the flight time, substitute this back into the horizontal motion equation:\[ range = x(t_{final}) = v_x \cdot t_{final} \]For our volleyball example, once the flight time is known, use it in the horizontal equation to find how far down the court the volleyball lands.
The range is affected by several factors:

• Initial Speed: The greater the initial speed, the larger the range, assuming the angle is optimized.
• Launch Angle: Typically, an angle of 45° yields the maximum range for a given speed in the absence of air resistance.

Understanding the range gives insights into how effectively a player can serve or spike a volleyball across the net.