Problem 40 - Thomas Calculus

Question

Hitting a baseball under a wind gust A baseball is hit when it is 2.5 ft above the ground. It leaves the bat with an initial velocity of 145 \(\mathrm{ft} / \mathrm{sec}\) at a launch angle of \(23^{\circ} .\) At the instant the ball is hit, an instantaneous gust of wind blows against the ball, adding a component of \(-14 \mathbf{i}(\mathrm{ft} / \mathrm{sec})\) to the ball's initial velocity. A \(15-\mathrm{ft}\) - high fence lies 300 ft from home plate in the direction of the flight. a. Find a vector equation for the path of the baseball. b. How high does the baseball go, and when does it reach maximum height? c. Find the range and flight time of the baseball, assuming that the ball is not caught. d. When is the baseball 20 ft high? How far (ground distance) is the baseball from home plate at that height? e. Has the batter hit a home run? Explain.

Step-by-Step Solution

Answer

The ball does not clear the fence at 300 ft, so it is not a home run.

Step 1: Calculate Initial Velocity Components

The initial velocity of the baseball is given as 145 ft/sec at an angle of \(23^{\circ}\). Without the wind, the components would be \(v_{x0} = 145 \cos(23^{\circ})\) and \(v_{y0} = 145 \sin(23^{\circ})\). The wind adds a velocity component \(-14\ \mathbf{i}\), affecting the horizontal component. Thus, the new initial velocities are: \(v_{x0} = 145 \cos(23^{\circ}) - 14\) and \(v_{y0} = 145 \sin(23^{\circ})\). Calculate these using a calculator.

Step 2: Write the Parametric Equations for Motion

The equations of motion for the ball are: \(x(t) = (v_{x0})t\) for horizontal motion where \(v_{x0}\) is corrected for wind, and \(y(t) = 2.5 + (v_{y0})t - \frac{1}{2}gt^2\) for vertical motion with initial height 2.5 ft and gravitational acceleration \(g = 32\ \mathrm{ft}/\mathrm{sec}^2\). Substitute the values obtained in Step 1 into these equations.

Step 3: Determine Maximum Height

To find when the ball reaches maximum height, differentiate the vertical equation \(y'(t)\) and set \(y'(t) = 0\). Solve for \(t\), then substitute this \(t\) back into \(y(t)\) to find the maximum height.

Step 4: Calculate Range and Flight Time

Solve the equation \(y(t) = 0\) using the quadratic formula to find when the ball hits the ground again, which gives the flight time. Use \(x(t)\) with this \(t\) to find the range by substituting \(t\) into the horizontal equation to get the distance covered.

Step 5: Find When Ball is 20 ft High

Solve the vertical equation \(y(t) = 20\) to find when the ball reaches 20 ft. This gives two times because the ball passes through 20 ft twice. Find corresponding \(x(t)\) values for each \(t\) to determine how far from home plate the ball is at each time.

Step 6: Determine if the Ball Clears the Fence

Use the equation \(x(t) = 300\) to determine \(t\) when the ball reaches the fence. Substitute this \(t\) into the vertical equation \(y(t)\) to see if the height is greater than 15 ft, the height of the fence.

Key Concepts

Vector Equations

Let's explore what a vector equation for projectile motion looks like. In this scenario, the path of the baseball can be modeled using a vector equation that describes its motion over time. The vector equation breaks down into two parts: the horizontal and vertical components.

• The **horizontal component** is affected by the wind and is calculated using the adjusted initial velocity, which incorporates the wind's effect.
• The **vertical component** is influenced by gravity and starts with an initial 2.5 ft due to the ball's starting height.

Together, these equations will help us comprehend the baseball's trajectory and answer questions about its maximum height, range, and whether or not it becomes a home run.

Initial Velocity Components

When analyzing projectile motion, it's crucial to break down initial velocity into horizontal and vertical components. These components help us understand and calculate the path of the baseball.

• **Horizontal Velocity Component**: This is given by the formula: \( v_{x0} = 145 \cos(23^{\circ}) - 14 \) where the cosine function calculates the horizontal component of the velocity vector, and the additional \(-14\) is due to the wind acting against the ball.
• **Vertical Velocity Component**: The vertical component is calculated as: \( v_{y0} = 145 \sin(23^{\circ}) \) indicating the vertical influence without any wind interference.

Understanding these components helps significantly when solving for other conditions such as maximum height and flight time.

Maximum Height Calculation

To determine how high an object will go in projectile motion, we need to investigate when its vertical velocity changes direction. This occurs when the derivative of the vertical position equation equals zero, indicating the peak point where the speed is entirely horizontal.

• **Find the Time of Maximum Height**: Differentiate the vertical motion equation \( y(t) = 2.5 + v_{y0}t - \frac{1}{2}gt^2 \) Set the derivative \( y'(t) \) to zero and solve for time \( t \).
• **Calculate Maximum Height**: Once \( t \) is found, substitute it back into \( y(t) \) to find the maximum height the baseball reaches.

This helps not just in sports but in any application where altitude maximization is relevant.

Range and Flight Time

Understanding the range and flight time of the projectile is essential in evaluating its entire path. In this task, finding when the ball hits the ground again gives the flight time, while the horizontal distance covered during this time is its range.

• **Flight Time Calculation**: Solve the vertical motion equation \( y(t) = 0 \) with the quadratic formula to determine when the projectile returns to the ground level.
• **Range Calculation**: Substitute the derived time into the horizontal motion equation \( x(t) = v_{x0}t \) to find how far the baseball travels.

These calculations are invaluable in sports analytics and engineering, showing how forces and angles impact motion.

Parametric Equations

The use of parametric equations allows each component of a projectile's path to be defined separately, offering a clearer way to analyze complex motion.

• **Horizontal Motion**: The equation \( x(t) = (v_{x0})t \) focuses on how far the ball travels over time, integrating the horizontal velocity component.
• **Vertical Motion**: Involves the equation \( y(t) = 2.5 + (v_{y0})t - \frac{1}{2}gt^2 \), capturing the loft and descent of the ball.

Parametric equations are a versatile tool used in fields ranging from animation to robotics, where defining precise paths in multidimensional spaces is necessary.