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A professional golfer hits her 5 -iron \(155 \mathrm{m}(170\) yd). A 46.0 -g golf ball experiences a drag force of magnitude \(R=C v^{2},\) and has a terminal speed of \(44.0 \mathrm{m} / \mathrm{s} .\) (a) Calculate the drag constant \(C\) for the golf ball. (b) Use a numerical method to calculate the trajectory of this shot. If the initial velocity of the ball makes an angle of \(31.0^{\circ}\) (the loft angle) with the horizontal, what initial speed must the ball have to reach the 155 -m distance? (c) If this same golfer hits her 9 -iron \(\left(47.0^{\circ} \text { loft }\right)\) a distance of \(119 \mathrm{m},\) what is the initial speed of the ball in this case? Discuss the differences in trajectories between the two shots.

Short Answer

Expert verified
The constant \(C\) for the golf ball can be calculated using the provided terminal speed and drag force. A numerical method involving equations of motion is applied to find the initial speeds that let the ball cover the given distances for specific angles. Due to high loft angle, the 9-iron shot would reach a higher peak but cover a smaller horizontal distance compared to the 5-iron shot.

Step by step solution

01

Calculation of Drag Constant

The drag force on the ball while it's falling freely at the terminal speed \(v_t\) is balanced by its weight. And the drag force formula is given by \(R = C v^2\). At terminal speed, \(R = mg\). Therefore, we have \(mg = C v_t^2\). Solving this equation for \(C\), the drag coefficient is thus \(C = \frac{mg}{v_t^2}\). Substitute \(m = 46.0\, g = 0.046\, kg\), \(g = 9.8\, m/s^2\), and \(v_t = 44.0\, m/s\) into the equation to get the value of \(C\).
02

Calculation of Initial Speed at a Given Angle

To calculate the trajectory and initial speed required, a numerical method such as a suitable computer algorithm can be used. One can use the Euler method or other similar ones. The motion of the projectile under air resistance is governed by the equations: \(dx/dt = v_x\), \(dy/dt = v_y\), \(dv_x/dt = -C v v_x/m\) and \(dv_y/dt = -g - C v v_y /m\), where \(x\) and \(y\) are the horizontal and vertical distances, \(v_x\) and \(v_y\) are the horizontal and vertical components of velocity respectively. The value of \(C\) obtained in the previous step, initial angles of 31 degrees and 47 degrees given in the problem and the equations of motion are used to find the initial speeds that make the projectile cover the stated distances.
03

Discussion on Differences in Trajectories

After calculating the initial speeds for both shots, we discuss the difference in their trajectories. This can be visualized with the help of a graph plotting \(x\) vs \(y\) for both cases. Due to the higher loft angle, the 9-iron shot would reach a higher peak height but would travel a smaller horizontal distance compared to the 5-iron shot.

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Key Concepts

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

Drag Force
When you think about a projectile moving through the air, one of the significant forces acting upon it is the drag force. This force opposes the projectile's motion and is influenced by factors including the object's speed and surface characteristics. In simplistic terms, drag can be likened to the sensation of wind resistance you feel when you stick your hand out of a moving car.

For our golf ball flying across the field, the drag force is calculated using the formula:
  • R = C \(v^2\): where \(R\) is the magnitude of the drag force, \(C\) is the drag constant, and \(v\) is the speed.
Determining the drag constant, \(C\), is a crucial step in understanding how the ball's motion will be influenced. For objects at terminal speed, the drag force equates to the object's weight, allowing us to solve for \(C\) using the equation:
  • C = \(\frac{mg}{v_t^2}\): Here, \(m\) stands for the mass of the object, \(g\) for gravitational acceleration, and \(v_t\) for terminal speed.
Terminal Speed
Terminal speed is achieved when an object moving through a fluid (like air) no longer accelerates. This happens when the drag force matches the object's weight. For example, your golf ball reaches its terminal speed when the push of gravity pulling it down is exactly countered by the upward drag force.

To calculate terminal speed, consider the balance of forces:
  • The formula for the drag force: \(R = C v^2\)
  • Balancing with weight: \(R = mg\), hence, \(mg = C v_t^2\)
Rearranging gives you the precise speed at which these forces balance:
  • v_t = \(\sqrt{\frac{mg}{C}}\)
This concept is pivotal in projectile motion, as it helps in understanding exactly how fast the projectile can go without accelerating further. The golfer's ball, for instance, has a terminal velocity of 44 m/s, which it won't exceed unless more force is applied.
Numerical Methods
Numerical methods are mathematical techniques often employed to solve complex equations that don’t have simple analytical solutions. When calculating projectile trajectories, these methods help us navigate the complexities introduced by factors like drag force.

One useful method is the Euler's method, a straightforward approach to finding approximate solutions. Euler’s method involves breaking time into small intervals and using calculus principles to approximate the state of the system at the next interval.

In projectile motion, the following differential equations are crucial:
  • Horizontal: \(\frac{dx}{dt} = v_x\)
  • Vertical: \(\frac{dy}{dt} = v_y\)
  • Horizontal velocity: \(\frac{dv_x}{dt} = -\frac{C v v_x}{m}\)
  • Vertical velocity: \(\frac{dv_y}{dt} = -g - \frac{C v v_y}{m}\)
Here, \(v_x\) and \(v_y\) are the velocity components in the horizontal and vertical directions, respectively. Understanding and applying these equations enables one to simulate the golf ball’s flight under real-world conditions.
Projectile Trajectory
Projectile trajectory is the path that a projectile follows under the influence of gravity and other forces. For our golfer, the trajectory depends heavily on the angle at which the ball is struck and the initial speed.

A projectile’s path is traditionally parabolic without air resistance, but when you introduce forces like drag, the path becomes more complex. For example, two golf shots with different angles (31° and 47°) demonstrate varied trajectories.

Key factors affecting trajectory include:
  • Initial speed: A higher speed increases the distance covered.
  • Launch angle: Angles such as 31° (with less loft) hit the ball further, while 47° (more loft) results in a higher but shorter trajectory.
  • External forces: The drag constant, calculated earlier, directly impacts the projectile's path.
Understanding these influences allows golfers, and students studying physics, to accurately predict and adjust their shots for desired outcomes. By simulating each scenario, you can compare and understand the differences in paths, making it easier to fine-tune skills or calculations.

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Most popular questions from this chapter

A student stands in an elevator that is continuously accelerating upward with acceleration a. Her backpack is sitting on the floor next to the wall. The width of the elevator car is \(I_{n}\) The student gives her backpack a quick kick at \(t=0\) imparting to it speed \(v\), and making it slide across the elevator floor. At time \(t,\) the backpack hits the opposite wall. Find the coefficient of kinetic friction \(\mu_{k}\) between the backpack and the elevator floor.

A sky diver of mass \(80.0 \mathrm{kg}\) jumps from a slow-moving aircraft and reaches a terminal speed of \(50.0 \mathrm{m} / \mathrm{s} .\) (a) What is the acceleration of the sky diver when her speed is \(30.0 \mathrm{m} / \mathrm{s} ?\) What is the drag force on the diver when her speed is (b) \(50.0 \mathrm{m} / \mathrm{s} ?\) (c) \(30.0 \mathrm{m} / \mathrm{s} ?\)

A light string can support a stationary hanging load of \(25.0 \mathrm{kg}\) before breaking. A \(3.00-\mathrm{kg}\) object attached to the string rotates on a horizontal, frictionless table in a circle of radius \(0.800 \mathrm{m},\) while the other end of the string is held fixed. What range of speeds can the object have before the string breaks?

An amusement park ride consists of a rotating circular platform \(8.00 \mathrm{m}\) in diameter from which \(10.0-\mathrm{kg}\) seats are suspended at the end of \(2.50-\mathrm{m}\) massless chains (Fig. \(P(6.63) .\) When the system rotates, the chains make an angle \(\theta=28.0^{\circ}\) with the vertical. (a) What is the speed of each seat? (b) Draw a free-body diagram of a 40.0 -kg child riding in a seat and find the tension in the chain.

The pilot of an airplane executes a constant-speed loop-theloop maneuver in a vertical circle. The speed of the airplane is \(300 \mathrm{mi} / \mathrm{h},\) and the radius of the circle is \(1200 \mathrm{ft}\). (a) What is the pilot's apparent weight at the lowest point if his true weight is 160 lb? (b) What is his apparent weight at the highest point? (c) What If? Describe how the pilot could experience weightlessness if both the radius and the speed can be varied. (Note: His apparent weight is equal to the magnitude of the force exerted by the seat on his body.)

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