Problem 96 - Physics for Scientists and Engineers with Modern Physics

Question

A batter hits a fly ball which leaves the bat \(0.90 \mathrm{~m}\) above the ground at an angle of \(61^{\circ}\) with an initial speed of \(28 \mathrm{~m} / \mathrm{s}\) heading toward centerfield. Ignore air resistance. \((a)\) How far from home plate would the ball land if not caught? \((b)\) The ball is caught by the centerfielder who, starting at a distance of \(105 \mathrm{~m}\) from home plate, runs straight toward home plate at a constant speed and makes the catch at ground level. Find his speed.

Step-by-Step Solution

Answer

(a) The ball lands approximately 68.55 m from home plate. (b) The centerfielder's speed is approximately 7.24 m/s.

Step 1: Break Down the Initial Velocity into Components

The initial velocity of the ball can be broken down into horizontal and vertical components using trigonometry. The horizontal component is given by \( v_{x_0} = v_0 \cos(\theta) \) and the vertical component is \( v_{y_0} = v_0 \sin(\theta) \). Substitute the given values: \( v_0 = 28 \mathrm{~m/s} \) and \( \theta = 61^{\circ} \).\[ v_{x_0} = 28 \cos(61^{\circ}) \approx 13.64 \mathrm{~m/s} \]\[ v_{y_0} = 28 \sin(61^{\circ}) \approx 24.30 \mathrm{~m/s} \]

Step 2: Determine Time of Flight

The time it takes for the ball to reach the ground can be found using the equation for vertical motion. Use the formula: \( y = y_0 + v_{y_0}t - \frac{1}{2}gt^2 \), where \( y_0 = 0.90 \mathrm{~m} \), \( y = 0 \) (ground level), and \( g = 9.81 \mathrm{~m/s^2} \).Substitute in the known values:\[ 0 = 0.90 + 24.30t - \frac{1}{2}(9.81)t^2 \] Solving this quadratic equation gives \( t \approx 5.03 \mathrm{~s} \).

Step 3: Calculate Horizontal Distance

The horizontal distance the ball travels is given by \( x = v_{x_0} t \). Use the horizontal velocity calculated in Step 1 and the time from Step 2:\[ x = 13.64 \times 5.03 \approx 68.55 \mathrm{~m} \]

Step 4: Calculate the Speed of the Centerfielder

The centerfielder starts 105 m from home plate and covers the remaining distance to catch the ball. The distance he needs to cover is \( 105 - 68.55 = 36.45 \mathrm{~m} \). Using the flight time of \( t = 5.03 \mathrm{~s} \), his speed \( v = \frac{\text{distance}}{\text{time}} \) is:\[ v = \frac{36.45}{5.03} \approx 7.24 \mathrm{~m/s} \]

Key Concepts

Kinematics

Kinematics is the branch of mechanics that describes the motion of objects without considering the forces causing such motion. In projectile motion problems like our fly ball scenario, we focus on key kinematic equations that relate displacement, velocity, and acceleration over time.
For projectiles, it's important to remember:

• Horizontal and vertical motions can be independently analyzed.
• Horizontal motion occurs at a constant velocity since there is no horizontal acceleration (ignoring air resistance).
• Vertical motion is influenced by gravitational acceleration, which is constant at approximately 9.81 m/s².

In our exercise, we are tasked with analyzing a fly ball's motion where initial parameters are given, and we employ kinematic formulas to predict the landing distance and evaluate the centerfielder's speed.

Trigonometric Decomposition

Trigonometric decomposition is a crucial step in solving projectile motion problems. It involves breaking down the initial velocity into horizontal and vertical components, making it easier to study them separately.
Given an initial velocity, \(v_0\), and launch angle, \(\theta\), use trigonometry:

• The horizontal component: \(v_{x_0} = v_0 \cos(\theta)\)
• The vertical component: \(v_{y_0} = v_0 \sin(\theta)\)

This not only simplifies calculations but allows us to apply the relevant kinematic equations correctly. In the example, the initial velocity of 28 m/s at an angle of 61° is decomposed into approximately 13.64 m/s horizontally and 24.30 m/s vertically. This tells us how fast the ball is moving in each direction right after being struck.

Quadratic Equation

Solving the quadratic equation is a common task when analyzing projectile motion, especially when determining the time of flight or the vertical motion of a projectile. In our exercise, we use the equation:\[ y = y_0 + v_{y_0}t - \frac{1}{2}gt^2 \]This equation describes how the vertical position changes over time, taking into account initial height \(y_0\), initial vertical velocity \(v_{y_0}\), and gravitational acceleration \(g\).
Rearranging and solving for \(t\) typically involves using the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Where \(a = -\frac{1}{2}g\), \(b = v_{y_0}\), and \(c = y_0\). In our ball scenario, solving this gives us the time, approximately 5.03 seconds, indicating how long the projectile stays in the air before hitting the ground.