91Ó°ÊÓ

A batter hits the baseball \(A\) with an initial velocity of \(v_{0}=100 \mathrm{ft} / \mathrm{sec}\) directly toward fielder \(B\) at an angle of \(30^{\circ}\) to the horizontal; the initial position of the ball is \(3 \mathrm{ft}\) above ground level. Fielder \(B\) requires \(\frac{1}{4}\) sec to judge where the ball should be caught and begins moving to that position with constant speed. Because of great experience, fielder \(B\) chooses his running speed so that he arrives at the "catch position" simultaneously with the baseball. The catch position is the field location at which the ball altitude is \(7 \mathrm{ft}\). Determine the velocity of the ball relative to the fielder at the instant the catch is made.

Step by step solution

01

Determine the Ball’s Trajectory

First, we need to find the equations of motion for the baseball's horizontal and vertical positions as functions of time. Using the initial speed of 100 ft/sec and the launch angle of 30 degrees, we decompose the velocity into horizontal and vertical components:- Horizontal velocity: \( v_{0x} = v_{0} \cdot \cos(30^{\circ}) = 100 \cdot \frac{\sqrt{3}}{2} = 50\sqrt{3} \text{ ft/sec} \)- Vertical velocity: \( v_{0y} = v_{0} \cdot \sin(30^{\circ}) = 100 \cdot \frac{1}{2} = 50 \text{ ft/sec} \)The equations of motion, considering gravity \(g = 32 \text{ ft/sec}^2\), are:-horizontal position: \( x(t) = v_{0x} \cdot t \)-vertical position: \( y(t) = 3 + v_{0y} \cdot t - \frac{1}{2} g t^2 \)

02

Solve for Time to Reach 7 ft Altitude

To find the time when the ball is at a 7 ft altitude, we use the vertical position equation:\[ 7 = 3 + 50t - 16t^2 \]Simplifying, we get:\[ 16t^2 - 50t + 4 = 0 \]Solving this quadratic equation using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a = 16\), \(b = -50\), \(c = 4\), we find:\[ t = \frac{50 \pm \sqrt{(-50)^2 - 4 \cdot 16 \cdot 4}}{2 \cdot 16} \]\[ t = \frac{50 \pm \sqrt{2500 - 256}}{32} \]\[ t = \frac{50 \pm \sqrt{2244}}{32} \]\[ t \approx 3.062 \, ext{seconds} \] (taking the positive root as time cannot be negative)

03

Determine Horizontal Distance Covered

Using the time found, calculate how far field position B is by substituting \(t\) into the horizontal position equation:\[ x(t) = 50\sqrt{3} \times 3.062 \approx 265.97 \, \text{ft} \]This means the fielder needs to move to the position at approximately 266 ft away from the starting point in the horizontal direction.

04

Calculate Fielder's Running Speed

The fielder has 2.812 seconds to move (since he starts after 0.25 seconds). The distance he needs to cover is approximately 266 feet.Running speed: divide distance by time:\[ \text{Speed} = \frac{265.97}{3.062 - 0.25} \approx \frac{265.97}{2.812} \approx 94.59 \, \text{ft/sec} \]

05

Calculate Relative Velocity of the Ball to the Fielder

The velocity of the ball relative to the fielder is given by the vector difference between their velocities.- Horizontal velocity of ball relative to fielder: \( 50\sqrt{3} - 94.59 \)- The vertical velocity of the ball (at the catch): \( v_{y} = 50 - 32 \times 3.062 \approx -48.0 \, \text{ft/sec} \)The magnitude of relative velocity is then:\[ \text{Relative Velocity} = \sqrt{(50\sqrt{3} - 94.59)^2 + (-48)^2} \]Computing, \( 50\sqrt{3} \approx 86.6 \), so:\[ \sqrt{(86.6 - 94.59)^2 + (-48)^2} = \sqrt{(-7.99)^2 + 2304} \approx 50 \, \text{ft/sec} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Relative Velocity

When we talk about relative velocity, we essentially mean the speed and direction of one object as observed from another moving object. In this exercise, we are comparing the velocity of the baseball to the velocity of the fielder.

To determine the relative velocity of the ball to the fielder, we need to understand several components:

• The ball's velocity - This is derived from its initial speed and the effect of gravity over time.
• The fielder's velocity - This is constant and directed towards the point where he plans to catch the ball.

To find the relative velocity, we subtract the fielder's velocity vector from the ball's velocity vector.
The key takeaway is that relative velocity helps us understand motion in a dynamic framework, by comparing the speeds of different objects from a particular reference point.

Equations of Motion

The equations of motion describe how objects move in different circumstances of speed, direction, and force. For projectile motion, we decompose the movement into two directions: horizontal and vertical.

In the horizontal direction, there’s no acceleration (ignoring air resistance). This means the horizontal velocity remains constant throughout the flight. The formula for horizontal motion is:

• \( x(t) = v_{0x} \times t \)

In the vertical direction, gravity acts to change the velocity of the object. Hence, the vertical motion includes acceleration, which is downward due to gravity. Its formula includes initial velocity, time, and the gravitational constant:

• \( y(t) = y_0 + v_{0y} \times t - \frac{1}{2} g t^2 \)

Where \( y_0 \) is the initial vertical position.

Quadratic Equation

Quadratic equations arise naturally in many physics problems, especially those involving projectile motion. In this problem, the need to calculate the time when the ball reaches the height of 7 feet requires solving a quadratic equation.

The equation we end up with is:

• \( 16t^2 - 50t + 4 = 0 \)

We use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which helps us find the possible "t" values representing time.

The quadratic formula breaks down as follows:

• \( a = 16 \)
• \( b = -50 \)
• \( c = 4 \)

By solving this, we discover the time required for the ball to ascend and then descend to the desired height, shedding light on periods of ascent and descent in projectile motion.

Horizontal and Vertical Components

The concept of breaking down a projectile's initial velocity into horizontal and vertical components is fundamental in analyzing its trajectory.

At launch, the baseball's velocity can be split into:

• Horizontal component (\( v_{0x} \)): This is calculated by \( v_{0} \cos(\theta) \).
• Vertical component (\( v_{0y} \)): This is calculated by \( v_{0} \sin(\theta) \).

In this exercise, the launch angle \( \theta \) is \( 30^\circ \). Substituting the values gives:

• \( v_{0x} = 100 \times \cos(30^\circ) = 50\sqrt{3} \text{ ft/sec} \)
• \( v_{0y} = 100 \times \sin(30^\circ) = 50 \text{ ft/sec} \)

Understanding these components allows us to better predict the path the projectile will take and when it will hit certain altitudes, which is crucial for timing the catch in this scenario.