Problem 24 - Essential Calculus Early Transcendentals

Question

A batter hits a baseball 3 ft above the ground toward the center field fence, which is 10 ft high and 400 ft from home plate. The ball leaves the bat with speed 115 \(\mathrm{ft} / \mathrm{s}\) at an angle \(50^{\circ}\) above the horizontal. Is it a home run? (In other words, does the ball clear the fence?)

Step-by-Step Solution

Answer

Yes, the ball is a home run as it clears the fence's height.

Step 1: Break Down Initial Parameters

The initial height of the ball is given as 3 ft. The fence is 400 ft away and 10 ft tall. The initial speed of the ball is 115 ft/s at an angle of 50° above the horizontal.

Step 2: Calculate Horizontal and Vertical Velocity Components

We decompose the initial velocity into horizontal and vertical components using the given angle.\[v_{x} = 115 \cos(50^{\circ}) \approx 73.94 \text{ ft/s}\]\[v_{y} = 115 \sin(50^{\circ}) \approx 88.06 \text{ ft/s}\]

Step 3: Calculate Time to Reach the Fence

Using the horizontal velocity component, calculate the time it takes to reach the fence:\[t = \frac{\text{distance to fence}}{v_{x}} = \frac{400}{73.94} \approx 5.41 \text{ seconds}\]

Step 4: Calculate Vertical Position at the Fence

Using the time calculated, compute the vertical position of the ball at the moment it reaches the fence:\[y(t) = 3 + v_{y} t - \frac{1}{2} g t^2\]where \(g = 32 \text{ ft/s}^2\) (acceleration due to gravity).Plugging in the values:\[y(5.41) = 3 + 88.06 \times 5.41 - \frac{1}{2} \times 32 \times (5.41)^2 \approx 14.23 \text{ ft}\]

Step 5: Determine if the Ball Clears the Fence

Compare the calculated height of the ball at the fence with the height of the fence. Since 14.23 ft > 10 ft, the ball clears the fence.

Key Concepts

Horizontal Velocity

In projectile motion, horizontal velocity refers to the component of an object's velocity that acts parallel to the ground. When a baseball is hit, its speed can be broken up into horizontal and vertical components. Although its vertical speed is subject to change due to gravity, the horizontal velocity remains constant throughout the flight of the projectile. This stability occurs because, in an ideal system without air resistance, there are no forces applied in the horizontal direction. To calculate the horizontal component, we use the initial speed and the angle:

• This calculation uses the cosine function:
  \(v_{x} = \text{Initial Speed} \times \cos(\text{Angle}) \)
• For our exercise, the horizontal velocity, \(v_x\), is \( 73.94 \, \text{ft/s} \).

Understanding the horizontal velocity component is crucial when predicting how long the projectile will travel before reaching the location of interest, in this case, the fence. In the solution provided, it determines the time the ball takes to travel the distance to the fence, helping to ascertain whether the ball clears the fence or not.

Vertical Velocity

Vertical velocity is a key factor in analyzing projectile motion, particularly with objects thrown or hit with an initial angle above or below the horizontal. The vertical component of velocity indicates how fast an object is moving upwards or downwards at any given point in its trajectory. The initial vertical velocity can be influenced by both the speed and angle at which an object is projected. For our baseball example, the formula used is:

• The vertical component is calculated using the sine of the angle:
  \(v_{y} = \text{Initial Speed} \times \sin(\text{Angle}) \)
• In the problem, the initial vertical velocity, \(v_y\), turns out to be \( 88.06 \, \text{ft/s} \).

Throughout its flight, the vertical component changes due to gravity, which continuously acts downward. This constant gravitational influence causes the baseball to rise to a peak, slow down, and eventually start descending back toward the ground. Calculating the height of the ball at a specific moment — such as when it reaches the fence — involves understanding the impact of both its initial vertical velocity and the deceleration due to gravity.

Acceleration Due to Gravity

Gravity plays a fundamental role in projectile motion by affecting only the vertical component of an object's velocity. Acceleration due to gravity is a consistent force that pulls objects downward towards the earth. For most practical activities on Earth's surface, the acceleration due to gravity is approximately \(32 \, \text{ft/s}^2 \). This factor is crucial for calculating how the vertical velocity and position of any projectile change over time. As the baseball travels through its trajectory:

• Gravity slows its upward movement until it reaches the apex or highest point.
• Once at its peak, gravity accelerates the ball downward, increasing its fall speed.

To determine the height of the ball when it reaches the fence, the gravitational force's effect must be integrated into the equation used to calculate vertical position. The equation:
\(y(t) = \text{Initial height} + v_y \times t - \frac{1}{2} g t^2\) demonstrates how height is calculated by considering both initial upward momentum and the pull of gravity. Grasping the concept of gravity's influence allows us to predict if a trajectory high enough ensures a successful home run by clearing a fence.