/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 136 A baseball is hit at Fenway Park... [FREE SOLUTION] | 91Ó°ÊÓ

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A baseball is hit at Fenway Park in Boston at a point \(0.762 \mathrm{~m}\) above home plate with an initial velocity of \(33.53 \mathrm{~m} / \mathrm{s}\) directed \(55.0^{\circ}\) above the horizontal. The ball is observed to clear the 11.28 -m-high wall in left field (known as the "green monster") \(5.00 \mathrm{~s}\) after it is hit, at a point just inside the left-field foulline pole. Find (a) the horizontal distance down the left-field foul line from home plate to the wall; (b) the vertical distance by which the ball clears the wall; (c) the horizontal and vertical displacements of the ball with respect to home plate \(0.500 \mathrm{~s}\) before it clears the wall.

Short Answer

Expert verified
(a) 96.25 m; (b) 4.4325 m; (c) Horizontal: 9.625 m, Vertical: 1.13625 m.

Step by step solution

01

Identify Initial Conditions

The initial height of the ball is given as \( y_0 = 0.762 \, \mathrm{m} \). The initial velocity is \( v_0 = 33.53 \, \mathrm{m/s} \) and the launch angle is \( \theta = 55.0^\circ \). The gravitational acceleration \( g = 9.81 \, \mathrm{m/s}^2 \) is acting downwards.
02

Resolve the Initial Velocity into Components

Calculate the horizontal and vertical components of the initial velocity using trigonometric functions:- Horizontal: \( v_{0x} = v_0 \cdot \cos(\theta) = 33.53 \cdot \cos(55.0^\circ) \approx 19.25 \, \mathrm{m/s} \)- Vertical: \( v_{0y} = v_0 \cdot \sin(\theta) = 33.53 \cdot \sin(55.0^\circ) \approx 27.50 \, \mathrm{m/s} \).
03

Calculate Horizontal Distance to the Wall

Using the time \( t = 5.00 \, \mathrm{s} \), calculate the horizontal distance with the formula:\[ x = v_{0x} \cdot t = 19.25 \cdot 5.00 = 96.25 \, \mathrm{m} \].
04

Calculate Final Vertical Position After 5 Seconds

The general equation for vertical motion is \( y = y_0 + v_{0y} t - \frac{1}{2} g t^2 \). Substitute given values:\[ y = 0.762 + 27.50 \cdot 5.00 - \frac{1}{2} \cdot 9.81 \cdot (5.00)^2 \approx 15.7125 \, \mathrm{m} \].
05

Calculate Vertical Clearance Over the Wall

Subtract the height of the wall from the final vertical position:\[ \text{Clearance} = 15.7125 - 11.28 = 4.4325 \, \mathrm{m} \].
06

Calculate Displacement 0.5 Seconds Before Clearing the Wall

Calculate position at \( t = 4.5 \, \mathrm{s} \):- Horizontal displacement: \( x_{4.5} = v_{0x} \cdot 4.5 = 19.25 \cdot 4.5 = 86.625 \, \mathrm{m} \).- Vertical position:\[ y_{4.5} = 0.762 + 27.50 \cdot 4.5 - \frac{1}{2} \cdot 9.81 \cdot (4.5)^2 \approx 14.57625 \, \mathrm{m} \].- Horizontal displacement before clearing: \( \Delta x = 96.25 - 86.625 = 9.625 \, \mathrm{m} \).- Vertical displacement before clearing: \( \Delta y = 15.7125 - 14.57625 = 1.13625 \, \mathrm{m} \).

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that describes the motion of objects without considering the forces that cause this motion. In projectile motion, kinematics is used to understand the trajectory of an object launched into the air. This involves assessing the object's initial velocity, the angle of projection, and the distances traveled both vertically and horizontally.

When dealing with a projectile like a baseball, initial velocity is crucial. It defines the speed and direction at which the baseball is launched. The kinematics equations involve calculations for both horizontal and vertical motions, allowing us to predict the motion across time. The equations you frequently use include determining displacement, velocity, and acceleration for an object in motion.

In our exercise, kinematics helps us calculate the baseball's trajectory. This includes determining how far the baseball travels until it clears a wall and how high it rises before descending. Understanding kinematics provides the foundation for solving these problems by breaking down the motion into simpler, manageable parts.
Trigonometry
Trigonometry plays a significant role in analyzing projectile motion, as it helps break down the components of a projectile's velocity. In the case of the baseball hit at Fenway Park, applying trigonometry allows us to resolve the initial velocity into horizontal and vertical components.

Using the angle of launch, trigonometric functions like sine and cosine help split the initial velocity. The horizontal component, often labeled as \( v_{0x} \), is calculated using cosine:
  • \( v_{0x} = v_0 \cdot \cos(\theta) \)
The vertical component, \( v_{0y} \), uses the sine function:
  • \( v_{0y} = v_0 \cdot \sin(\theta) \)
These calculations are crucial as they determine how far and how high the projectile will travel. Deploying trigonometry allows physics students to manage complex trajectories by simplifying the motion into directional components, each affected differently by gravity and other forces.
Gravitational acceleration
Gravitational acceleration is a constant that acts on all objects in projectile motion. On Earth, this acceleration approximates to \( 9.81 \, \mathrm{m/s}^2 \), pulling objects downward towards the Earth's surface. This gravitational force significantly influences the vertical component of a projectile's motion.

In our exercise, gravity affects the vertical component of the baseball's motion, constantly decelerating the upward velocity until the projectile reaches its peak height. Upon reaching this point, gravity accelerates the baseball downwards, increasing its velocity until it clears the wall or hits the ground. The standard kinematic equation used to describe this vertical motion is:
  • \( y = y_0 + v_{0y} t - \frac{1}{2} g t^2 \)
This formula includes gravitational acceleration \( g \) and helps predict the vertical position of the baseball at various points in time.

Thus, understanding gravitational acceleration is essential to solve projectile problems accurately. It ensures that each calculation for the trajectory in the vertical direction accounts for the constant pull downwards, which shapes the overall path of the projectile.

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