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Ch 08: Dynamics II: Motion in a Plane
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 8, Problem 68b

In the absence of air resistance, a projectile that lands at the elevation from which it was launched achieves maximum range when launched at a 45° angle. Suppose a projectile of mass m is launched with speed into a headwind that exerts a constant, horizontal retarding force Fwind=Fwindı^.\(\vec{F}\)_{\(\text{wind}\)} = -F_{\(\text{wind}\)} \(\hat{\imath}\). By what percentage is the maximum range of a 0.50 kg ball reduced if Fwind=0.60 NF_{\(\text{wind}\)}=0.60\(\text{ N}\)?

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Step 1: Begin by understanding the problem. The projectile is launched at an angle of 45° with an initial speed, and there is a horizontal retarding force due to the headwind. The goal is to determine the percentage reduction in the maximum range caused by this force.
Step 2: Write down the equations of motion for the projectile. The horizontal motion is affected by the retarding force, while the vertical motion is governed by gravity. The horizontal acceleration due to the headwind is given by \( a_x = \frac{-F_{\text{wind}}}{m} \), where \( F_{\text{wind}} \) is the force exerted by the wind and \( m \) is the mass of the projectile.
Step 3: Calculate the time of flight of the projectile. The time of flight is determined by the vertical motion. Using the kinematic equation \( t = \frac{2v_0 \sin \theta}{g} \), where \( v_0 \) is the initial speed, \( \theta \) is the launch angle, and \( g \) is the acceleration due to gravity.
Step 4: Determine the horizontal range of the projectile. The range is given by \( R = v_0 \cos \theta \cdot t + \frac{1}{2} a_x t^2 \), where \( a_x \) is the horizontal acceleration due to the wind. Substitute \( a_x = \frac{-F_{\text{wind}}}{m} \) and \( t \) from Step 3 into this equation.
Step 5: Compare the range with and without the wind. The range without the wind is \( R_0 = \frac{v_0^2 \sin 2\theta}{g} \). Calculate the percentage reduction in range using \( \text{Percentage Reduction} = \frac{R_0 - R}{R_0} \times 100 \). Substitute the given values \( m = 0.50 \text{ kg} \), \( F_{\text{wind}} = 0.60 \text{ N} \), and solve symbolically to find the percentage reduction.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Projectile Motion

Projectile motion refers to the motion of an object that is launched into the air and is subject to gravitational force. The trajectory of a projectile is typically parabolic, and its range is influenced by the launch angle, initial speed, and the effects of forces such as gravity and air resistance. In ideal conditions, the maximum range occurs at a 45° launch angle, balancing vertical and horizontal components of motion.
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Forces Acting on the Projectile

In this scenario, the projectile is affected by two main forces: gravity, which acts downward, and a horizontal retarding force due to the headwind. The retarding force reduces the horizontal component of the projectile's velocity, thereby affecting its range. Understanding how these forces interact is crucial for calculating the new range of the projectile when subjected to the headwind.
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Range Reduction Calculation

To determine the percentage reduction in range due to the headwind, one must first calculate the original range without the wind and then the new range with the wind's effect. The percentage reduction can be found using the formula: ((original range - new range) / original range) * 100%. This calculation highlights the impact of external forces on projectile motion and is essential for solving the given problem.
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Related Practice
Textbook Question

A 500 g steel block rotates on a steel table while attached to a 1.2-m-long hollow tube as shown in FIGURE CP8.70. Compressed air fed through the tube and ejected from a nozzle on the back of the block exerts a thrust force of 4.0 N perpendicular to the tube. The maximum tension the tube can withstand without breaking is 50 N. If the block starts from rest, how many revolutions does it make before the tube breaks?

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Textbook Question

2.0 kg ball swings in a vertical circle on the end of an 80-cm-long string. The tension in the string is 20 N when its angle from the highest point on the circle is θ = 30°. What is the ball's speed when θ = 30°?

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Textbook Question

The 10 mg bead in FIGURE CP8.69 is free to slide on a frictionless wire loop. The loop rotates about a vertical axis with angular velocity ω. If ω is less than some critical value ω꜀, the bead sits at the bottom of the spinning loop. When ω > ω꜀, the bead moves out to some angle θ. What is ω꜀ in rpm for the loop shown in the figure?

Textbook Question

For safety, elevators have a rotational governor, a device that is attached to and rotates with one of the elevator's pulleys. The governor, shown in FIGURE P8.63, is a disk with two hollow channels holding springs with metal blocks of mass m attached to their free ends. The faster the governor spins, the more the springs stretch. At a critical angular velocity ωc, the metal blocks contact the housing, which completes a circuit and activates an emergency brake. The spring force on a mass, which we will explore more thoroughly in Chapter 9, is FSp = k(r - L), where k is the spring constant measured in N/m, and L is the relaxed (unstretched) length of the spring. Suppose a rotational governor has L = 0.80R and the emergency brake activates when the metal blocks reach r = R. What is the critical angular velocity in rpm if R = 15cm, k = 20 N/m, and m = 25g? Ignore gravity.

Textbook Question

If a vertical cylinder of water (or any other liquid) rotates about its axis, as shown in FIGURE CP8.72, the surface forms a smooth curve. Assuming that the water rotates as a unit (i.e., all the water rotates with the same angular velocity), show that the shape of the surface is a parabola described by the equation z = (ω2 / 2g) r2. Hint: Each particle of water on the surface is subject to only two forces: gravity and the normal force due to the water underneath it. The normal force, as always, acts perpendicular to the surface.

Textbook Question

In the absence of air resistance, a projectile that lands at the elevation from which it was launched achieves maximum range when launched at a 45° angle. Suppose a projectile of mass m is launched with speed into a headwind that exerts a constant, horizontal retarding force Fwind=Fwindı^.\(\vec{F}\)_{\(\text{wind}\)} = -F_{\(\text{wind}\)} \(\hat{\imath}\).. Find an expression for the angle at which the range is maximum.

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