Problem 26 - Thomas Calculus

Question

Hitting a baseball under a wind gust A baseball is hit when it is 2.5 \(\mathrm{ft}\) 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 -ft-high fence lies 300 \(\mathrm{ft}\) 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 \(\mathrm{ft}\) 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 baseball reaches a maximum height of 54.3 ft in 2.0 sec, covers a range of 460 ft in 6.32 sec, is at 20 ft height twice at 144 ft and 312 ft from home, and clears the fence for a home run.

Step 1: Determine the initial velocity vector

Calculate the initial velocity vector of the baseball. The initial velocity given is 145 ft/sec at a launch angle of \(23^\circ\). The wind adds a component of \(-14 \mathbf{i}\). Thus, the initial velocity vector is: \( \mathbf{v_0} = (145 \cos(23^\circ) - 14) \mathbf{i} + (145 \sin(23^\circ)) \mathbf{j} \). Calculate the components separately:- Horizontal component: \( u_x = 145 \cos(23^\circ) - 14 \)- Vertical component: \( u_y = 145 \sin(23^\circ) \).

Step 2: Write the vector equation

The position vector \( \mathbf{r}(t) \) of the baseball at time \( t \) can be given by the parametric equations:\[ \mathbf{r}(t) = (u_x t) \mathbf{i} + (2.5 + u_y t - \frac{1}{2} g t^2) \mathbf{j} \]where \( g = 32 \text{ ft/sec}^2 \) is the acceleration due to gravity. Substitute \( u_x \) and \( u_y \) from Step 1 to get the full equation.

Step 3: Calculate maximum height and time to reach it

The vertical component of the velocity will be zero at the maximum height. Use \( v_y = u_y - g t \) and set it to zero: \[ 0 = u_y - 32t \]Solve for \( t \) to find the time to reach maximum height. Then, substitute \( t \) back into the height equation \( y(t) = 2.5 + u_y t - \frac{1}{2} \cdot 32 \cdot t^2 \) to find the maximum height.

Step 4: Determine range and flight time

The time of flight occurs when the vertical displacement \( y(t) = 0 \). Solve the equation:\[ 0 = 2.5 + u_y t - \frac{1}{2} \cdot 32 \cdot t^2 \]This is a quadratic equation in \( t \). Use the quadratic formula to find the time of flight. The range \( x(t) \) is calculated by substituting the flight time back into \( x(t) = u_x t \).

Step 5: Calculate the height of the baseball at 20 ft and ground distance

Substitute \( y(t) = 20 \) into the height equation to solve for \( t \).\[ 20 = 2.5 + u_y t - \frac{1}{2} \cdot 32 \cdot t^2 \]Solve the resulting quadratic equation twice to find the two possible \( t \) values. For each \( t \), calculate the horizontal distance \( x(t) = u_x t \) to find out how far the baseball is from the home plate.

Step 6: Determine if the baseball clears the fence

At the distance of the fence (300 ft), calculate the height of the baseball using the equation for \( x(t) = 300 \). Solve for \( t \), then plug \( t \) into the height equation:\[ y(t) = 2.5 + u_y t - \frac{1}{2} \cdot 32 \cdot t^2 \]If \( y(t) > 15 \) ft, the baseball clears the fence, otherwise, it doesn't.

Key Concepts

Baseball Trajectory

A baseball trajectory is an example of projectile motion, which describes the path that an object follows once it is propelled near the surface of the Earth. The baseball in this context follows a parabolic path due to gravity influencing its vertical motion. The starting point in analyzing baseball trajectory is to understand the initial conditions—this includes the initial height, launch angle, and velocity. In the given exercise, the ball is hit 2.5 feet above the ground and at a launch angle of 23 degrees with a velocity of 145 ft/sec. Additionally, the effect of a wind gust is added, demonstrating how external factors can alter the trajectory. To determine how the ball moves, we can use parametric equations that break down the motion into horizontal and vertical components. This involves calculating the initial horizontal and vertical velocities, taking into account the effect of the wind. Understanding these initial conditions allows us to predict various features of the trajectory, such as maximum height, range, and whether the ball will clear obstacles like fences.

Vectors in Physics

In physics, vectors are essential for describing quantities that have both magnitude and direction. Analyzing the flight of a baseball requires understanding vectors, as they succinctly describe the baseball's motion. In our exercise, the initial velocity of the baseball is a vector that can be decomposed into a horizontal component (along the x-axis) and a vertical component (along the y-axis).

• Horizontal component: influenced by initial speed and the angle of launch, as well as the impact of wind resistance.
• Vertical component: influenced by gravity and the initial upward motion.

These components can be expressed in vector form: \[ v_0 = (145 \cos(23^\circ) - 14) \mathbf{i} + (145 \sin(23^\circ)) \mathbf{j} \]Here, \( \mathbf{i} \) represents the horizontal direction, and \( \mathbf{j} \) the vertical. The versatility of vectors allows us to solve complex problems within projectile motion by breaking them down into simpler, manageable parts. This makes vectors a fundamental tool in understanding physics problems like the baseball trajectory.

Kinematics

Kinematics is the branch of physics that deals with the motion of objects, without considering the forces that cause the motion. To describe the baseball's motion, we employ kinematic equations that connect displacement, velocity, acceleration, and time. For our baseball example, the path is determined by the kinematic equations of motion, which involve:

• Initial velocity (velocity vector components)
• Acceleration due to gravity (which is \( -32 \text{ ft/sec}^2 \) here)
• Time of flight

These factors combine in the equation \[ \mathbf{r}(t) = (u_x t) \mathbf{i} + (2.5 + u_y t - \frac{1}{2} g t^2) \mathbf{j} \] to describe the position of the baseball at any time \( t \). By analyzing these equations, you can predict when the baseball reaches its peak—where its vertical velocity is zero—and when it returns to the initial height. Understanding kinematics is key to solving problems about flight time, reaching certain heights, and maximum range in projectile scenarios like baseball trajectories.

Quadratic Equations

Quadratic equations frequently arise in physics and engineering, especially when dealing with projectile motion. The structure of a quadratic equation is \[ ax^2 + bx + c = 0 \]. In the context of our baseball problem, solving for time involves quadratic equations because the equations of motion incorporate acceleration due to gravity.For example, to find when the baseball returns to the ground, or when it reaches certain heights, you set the vertical position equation \( y(t) = 2.5 + u_y t - \frac{1}{2} \cdot 32 \cdot t^2 \) to the desired height such as zero or 20 feet. Solving this yields a quadratic equation in \( t \).To solve for \( t \), we use the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \], which allows us to determine the possible times when these events occur. Learning to solve quadratic equations is crucial for predicting key points along the projectile's path, such as when it reaches maximum height or a certain distance.