/*! 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 45 When the effect of aerodynamic d... [FREE SOLUTION] | 91Ó°ÊÓ

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When the effect of aerodynamic drag is included, the \(y\) -acceleration of a baseball moving vertically upward is \(a_{u}=-g-k v^{2},\) while the acceleration when the ball is moving downward is \(a_{d}=-g+k v^{2},\) where \(k\) is a positive constant and \(v\) is the speed in meters per second. If the ball is thrown upward at \(30 \mathrm{m} / \mathrm{s}\) from essentially ground level, compute its maximum height \(h\) and its speed \(v_{f}\) upon impact with the ground. Take \(k\) to be \(0.006 \mathrm{m}^{-1}\) and assume that \(g\) is constant.

Short Answer

Expert verified
Maximum height is approximately 42.37 m, and the impact speed is about 30 m/s.

Step by step solution

01

Define Variables and Constants

First, let's define the given variables and constants. We have the initial speed \( v_0 = 30 \text{ m/s} \), the gravitational acceleration \( g = 9.81 \text{ m/s}^2 \), and the drag coefficient \( k = 0.006 \text{ m}^{-1} \). The upward acceleration is defined as \( a_u = -g - k v^2 \) and downward acceleration as \( a_d = -g + k v^2 \). Our goal is to find the maximum height \( h \) and the final speed upon impact \( v_f \).
02

Determine Maximum Height

At the maximum height, the velocity of the baseball is zero. Using the equation \( a_u = -g - k v^2 \) for vertical motion, integrate from \( v_0 \) to 0 to find the maximum height. The equation of motion is \(-g - kv^2 = v \frac{dv}{dy}\). Solving this differential equation from \( v_0 = 30 \text{ m/s} \) to \( v = 0 \), and integrating, gives the height \( h \).
03

Calculate Maximum Height

The integral \( \int{ \frac{v}{g + k v^2} dv } = \int{ dy } \) can be solved using substitution or a table of integrals. The result is \( y_{max} = \frac{1}{2k} \ln \left( 1 + \frac{k v_0^2}{g} \right) \). Substituting the known values for \( k \), \( v_0 \), and \( g \) gives \( h \approx 42.37 \text{ meters} \).
04

Solve for Final Speed on Impact

For the impact speed, we need to determine the velocity when the ball reaches ground level again using the equation \( a_d = -g + k v^2 \). From maximum height back to ground (\( y = 0 \)), the speed changes from zero upward back to a final speed downward. Use energy considerations or solve the integral \( \int{ \frac{v}{g - k v^2} dv } = \int{ -dy } \) from \( 0 \) to \( v_f \).
05

Calculate Final Impact Speed

The integration yields \( \int{ \frac{v}{g - k v^2} dv } = - \int{ dy } \), solved by using a partial fraction expansion or similar methods. The result for \( v_f \) is found to be approximately \( \sqrt{ \frac{2gh}{1 - k h } } \). Substituting \( h = 42.37 \text{ m} \), \( g = 9.81 \text{ m/s}^2 \), and \( k = 0.006 \text{ m}^{-1} \) gives \( v_f \approx 30 \text{ m/s} \).

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Key Concepts

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

Differential Equations
Differential equations play a crucial role in understanding motion when forces like aerodynamic drag are involved. They allow us to represent the relationships between an object's velocity, acceleration, and position over time. In this problem, two specific differential equations describe the baseball's motion.
  • Upward motion: \( a_u = -g - k v^2 \), where the drag force opposes gravity.
  • Downward motion: \( a_d = -g + k v^2 \), where the drag force combines with gravity.
These equations are expressions of Newton's second law, integrating both gravitational force and drag force, which depends on the square of velocity. Solving these differential equations either through integration or using known techniques allows us to predict how the baseball's speed changes during its flight. This understanding is essential for computing the maximum height the ball reaches and its speed upon impact with the ground.
Kinematics
Kinematics focuses on describing motion without regard to the forces causing it. In this exercise, kinematics helps determine how the baseball's position and velocity evolve over time. To find the maximum height the ball reaches, we consider the point when its upward velocity becomes zero. At this point, the forces of gravity and drag perfectly counterbalance the initial force imparted by throwing the ball. By integrating the velocity function derived from our differential equation, kinematics supplies us with the maximum height. Similarly, when analyzing the downward motion, kinematics describes how the initial velocity of zero transforms into a final speed when impacting the ground. This involves integrating the motion equations to relate height and velocity. By leveraging kinematic principles along with the differential equations, we can ascertain critical elements of the baseball's trajectory.
Integral Calculus
Integral calculus is fundamental in this problem, as it is used to solve differential equations and calculate the trajectory of the baseball. By integrating the differential equations that describe the baseball's acceleration, we find expressions for both height and velocity. For the upward motion, we perform an integral to find maximum height:\[ \int{ \frac{v}{g + k v^2} dv } = \int{ dy } \]Subsequently, after substituting the known values, we find an expression for the maximum height \( h \). Similarly, integral calculus is employed for calculating the final velocity as the baseball returns to ground level:\[ \int{ \frac{v}{g - k v^2} dv } = \int{ -dy } \]These integrals often require substitutions or use of integral tables to solve. Ultimately, integral calculus provides a robust method for progressing from acceleration to solve for position and velocity, effectively linking the kinematic description to real-world constants like aerodynamic drag.

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