Problem 36 - 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 \(-14 \mathrm{i}(\mathrm{ft} / \mathrm{sec})\) to the ball's initial velocity. \(\mathrm{A} 15\) -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 ft high? How far (ground distance) is e. Has the batter hit a home run? Explain.

Step-by-Step Solution

Answer

The baseball path is parametric, reaches max height 52.51 ft at 1.77 s. Range is 423.38 ft; clears 15 ft fence. It's a home run.

Step 1: Determine the Initial Velocity Components

We begin by calculating the initial velocity components for the baseball given the launch angle and the wind effect. The initial velocity of the baseball without wind is 145 ft/sec at an angle of 23°. - The horizontal component of the initial velocity is: \[ v_{0x} = 145\cos(23°) \approx 133.6 \, \text{ft/sec} \]- The vertical component of the initial velocity is: \[ v_{0y} = 145\sin(23°) \approx 56.5 \, \text{ft/sec} \]With the wind influence, the horizontal component modifies:\[ v_{0x} = 133.6 - 14 = 119.6 \, \text{ft/sec} \]

Step 2: Establish the Parametric Equations for Motion

To create a parametric form for the motion of the baseball, use the horizontal and vertical motion equations:- Horizontal (x) position as a function of time:\[ x(t) = v_{0x} t = 119.6t \]- Vertical (y) position as a function of time considering initial height (2.5 ft) and gravity:\[ y(t) = -16t^2 + v_{0y}t + 2.5 = -16t^2 + 56.5t + 2.5 \]

Step 3: Calculate Maximum Height and Time to Reach It

To find the maximum height, determine the time at which the vertical velocity becomes zero. The vertical velocity is given by:\[ v_y(t) = v_{0y} - gt = 56.5 - 32t \]Set \( v_y(t) = 0 \), therefore:\[ 56.5 - 32t = 0 \implies t = 1.77 \, \text{sec} \]Substitute back to find the maximum height:\[ y(1.77) = -16(1.77)^2 + 56.5(1.77) + 2.5 = 52.51 \, \text{ft} \]

Step 4: Determine Range and Flight Time

The range is the horizontal distance when the baseball hits the ground:Setting \( y(t)=0 \) in \( y(t) = -16t^2 + 56.5t + 2.5 \) gives us the flight time:Solve the quadratic:\[ -16t^2 + 56.5t + 2.5 = 0 \]Using the quadratic formula, \(t \approx 3.54 \, \text{sec}\).The range is:\[ x(3.54) = 119.6 \times 3.54 = 423.38 \, \text{ft} \]

Step 5: Verify When the Baseball is 20 ft High

Use the vertical position equation \( y(t) = -16t^2 + 56.5t + 2.5 \) to find when it is 20 ft high:\[ -16t^2 + 56.5t + 2.5 = 20 \]Solve the equation for both roots:\[ t \approx 0.42 \, \text{sec} \quad \text{and} \quad t \approx 3.19 \, \text{sec} \]Calculate the ground distance at 3.19 sec:\[ x(3.19) = 119.6 \times 3.19 = 381.53 \, \text{ft} \]

Step 6: Determine if It Clears the Fence (Home Run)

Check if the baseball clears a 15-ft high fence located at 300 ft from the origin:Find \( t \) when \( x(t) = 300 \):\[ 300 = 119.6t \implies t = 2.51 \, \text{sec} \]Evaluate height at this time:\[ y(2.51) = -16(2.51)^2 + 56.5(2.51) + 2.5 \approx 56.25 \, \text{ft} \]Since 56.25 ft > 15 ft, the baseball clears the fence.

Key Concepts

Vector Equations

Vector equations are a great way to represent the motion of objects like baseballs. In projectile motion, these equations specify how the object travels in a 2D space defined by both horizontal and vertical components.
For a baseball hit into the air, the initial velocity vector can initially be broken down into its horizontal and vertical components owing to the angle at which it is launched. When we add the wind component, the overall velocity is altered, especially the horizontal part.
The vector equation for the trajectory incorporates these aspects by utilizing both the modified horizontal velocity and the given vertical one. If we denote the influenced initial velocities by

• Horizontal component: \(v_x = 119.6 \text{ ft/sec}\)
• Vertical component: \(v_y = 56.5 \, \text{ft/sec}\)

Thus, they combine to form the motion path in vector form as \(\vec{r}(t) = (119.6t)\hat{i} + (-16t^2 + 56.5t + 2.5)\hat{j}\). Ultimately, this equation allows us to predict the baseball's position at any given time.

Initial Velocity Components

The initial velocity components are the foundation of analyzing projectile motion. These components originate from decomposing the initial velocity vector into horizontal and vertical parts.
To find these components, you need to use trigonometry based on the initial speed and the launch angle. For a velocity of 145 ft/sec at a launch angle of \(23^\circ\), we calculate:

• Horizontal component (\(v_{0x}\)) as \(145 \cos(23^\circ) = 133.6 \, \text{ft/sec}\)
• Vertical component (\(v_{0y}\)) as \(145 \sin(23^\circ) = 56.5 \, \text{ft/sec}\)

However, with a wind gust adding a component of \(-14 \, \text{ft/sec}\), the horizontal velocity becomes \(119.6 \, \text{ft/sec}\).
These components are essential as they define the projectile's motion equations and help determine how far and high the baseball will travel.

Parametric Equations

Parametric equations are useful for describing projectile motion by forming separate time-dependent equations for horizontal and vertical positions.
These equations handle the complexity of different forces acting simultaneously on an object, like the baseball in our problem tackled by gravity and wind.
For the baseball, the parametric equations considering its motion are:

• Horizontal position (x) as \(x(t) = 119.6t\), where \(119.6 \text{ ft/sec}\) is the adjusted initial horizontal velocity.
• Vertical position (y) as \(y(t) = -16t^2 + 56.5t + 2.5\), factoring in the gravitational acceleration of \(-32 \, \text{ft/sec}^2\) (half of it is \(-16\)) and the initial height (2.5 ft).

These parametric equations allow a deep understanding of both the flight path and trajectory of the baseball through time.

Maximum Height Calculation

Calculating the maximum height of a projectile involves identifying when the vertical velocity becomes zero, as this is the peak point in its path.
For the baseball, the vertical velocity at any time \(t\) is given by \(v_y(t) = 56.5 - 32t\). Setting it to zero to find the time at which maximum height occurs yields:

• \(56.5 - 32t = 0\)
• \(t = 1.77 \, \text{sec}\)

Plugging this time back into the vertical position equation \(y(t) = -16t^2 + 56.5t + 2.5\) provides the maximum height:

• \(y(1.77) = 52.51 \, \text{ft}\)

This calculation ensures that the highest point during the baseball's flight is accurately determined, offering insight into the baseball's performance concerning its launch parameters.