/*! 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 36 A baseball is struck with an ini... [FREE SOLUTION] | 91Ó°ÊÓ

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A baseball is struck with an initial speed of \(45 \mathrm{~m} \mathrm{~s}^{-1}\) (just over 100 \(\mathrm{mph}\) ) at an elevation angle of \(40^{\circ} .\) Find its path and compare this with the corresponding path when air resistance is neglected. [A baseball has mass \(0.30 \mathrm{~kg}\) and radius \(3.5 \mathrm{~cm}\). Assume the quadratic law of resistance.] Show that the equation of motion can be written in the form $$\frac{d v}{d t}=-g\left(k+\frac{v|v|}{V^{2}}\right)$$ where \(V\) is the terminal speed. Resolve this vector equation into two (coupled) scalar equations for \(v_{x}\) and \(v_{z}\) and perform a numerical solution. In this example, air resistance reduces the range by about \(35 \%\). It really is easier to hit a home run in Mile High stadium!

Short Answer

Expert verified
The path of the baseball when air resistance is considered is found to be significantly shorter due to the resistance faced by the ball. This path, when compared to the path without air resistance, is reduced by approximately 35%.

Step by step solution

01

Identify known quantities

Mass of baseball, \(m = 0.30 kg\), Radius of baseball, \(r = 3.5 cm\), Initial Speed, \(v = 45 m/s\), Angle of Projection, \(\theta = 40^{\circ}\)
02

Write the equation of motion

Write the equation of motion as given in the problem statement, \(\frac{d v}{d t}=-g\left(k+\frac{v|v|}{V^{2}}\right)\) where \(V\) is the terminal speed.
03

Resolve the equation into scalar components

Now using the angle, \(\theta\), resolve this vector equation into two (coupled) scalar equations for \(v_{x}\) and \(v_{z}\).
04

Solve numerically for \(v_{x}\) and \(v_{z}\)

Using standard numerical techniques solve the equations for \(v_{x}\) and \(v_{z}\) individually. This will give the different velocities as functions of time.
05

Find the path with air resistance

Path with air resistance can be found by integrating \(v_{x}\) and \(v_{z}\) with respect to time. This will give the x and z coordinates respectively. This will be compared with the ideal trajectory where air resistance is neglected.
06

Compare Paths

Compare the two paths. One is the path with air resistance and the other one being with no air resistance. As per the problem, the range will reduce by around 35% in presence of air resistance.

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Key Concepts

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

Air Resistance
Air resistance, also known as drag, plays a significant role in projectile motion, especially when dealing with high-speed objects like a baseball. When a baseball travels through the air, it encounters resistance that opposes its motion, thereby affecting its velocity and trajectory. The degree of this opposition depends on several factors, including:
  • The speed of the object: Higher speeds typically result in higher air resistance.
  • The cross-sectional area: Larger objects encounter more air resistance.
  • The density of the fluid (air in this case): Dense air provides more resistance.
  • The shape of the object, which influences how air flows around it.
Air resistance is often modeled using the quadratic law, which states that drag force is proportional to the square of the velocity: \[ F_d = -kv^2 \]where \(k\) is the drag coefficient. This introduces a non-linear component into the equations of motion. In our problem, air resistance significantly reduces the range of the baseball, cutting it by about 35% compared to when air resistance is neglected. This demonstrates why accounting for air resistance is essential in accurately predicting projectile paths.
Numerical Solution
To solve projectile motion problems involving air resistance, we often rely on numerical solutions. These solutions allow us to handle the complex dynamics that arise when the force of air resistance, which depends on velocity, is considered. Unlike simple projectile motion equations, which result in straightforward analytical solutions, the presence of air resistance requires solving coupled differential equations numerically.Here's what you might typically do:
  • Start by setting up the equations of motion. In our case, we have the equation: \(\frac{d v}{d t} = -g\left(k+\frac{v|v|}{V^{2}}\right)\)
  • Resolve this into horizontal and vertical components, \(v_x\) and \(v_z\).
  • Use numerical methods such as Euler's method or the Runge-Kutta method to approximate solutions for these velocity components over small time intervals.
Through numerical solutions, we can simulate and visualize the object's path, providing us with an understanding of the effect of air resistance on linear and angular projections. This approach is crucial when analytical methods fall short, ensuring that even complex real-world scenarios can be accurately modeled.
Terminal Velocity
Terminal velocity is a key concept in understanding how air resistance affects projectile motion. It refers to the maximum velocity an object can attain as it falls through a fluid, like air, when the force of drag equals the gravitational pull, resulting in a net force of zero. At this point, an object will continue to fall at a constant speed without further acceleration.For our baseball scenario:
  • As the baseball travels through the air, its velocity is initially very high, but air resistance increases with speed.
  • Eventually, the drag force becomes equal to the gravitational force, and the baseball stops accelerating, reaching terminal velocity: \( V = \sqrt{\frac{mg}{k}} \)where \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(k\) is the drag coefficient.
  • Under terminal velocity, the motion remains steady, influencing how far and fast the baseball can travel.
In projectile motion, terminal velocity serves as a boundary condition for realistic modeling, especially for longer durations in the air. It underlines the crucial balance between forces and highlights the limits for the realistic maximum speed attainable during the flight of objects.

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Most popular questions from this chapter

A particle of mass \(m\) can move on a rough horizontal table and is attached to a fixed point on the table by a light inextensible string of length \(b\). The resistance force exerted on the particle is \(-m K v\), where \(v\) is the velocity of the particle. Initially the string is taut and the particle is projected horizontally, at right angles to the string, with speed \(u\). Find the angle turned through by the string before the particle comes to rest. Find also the tension in the string at time \(t\).

A projectile is fired from the top of a conical mound of height \(h\) and base radius \(a\). What is the least projection speed that will allow the projectile to clear the mound? [Hint. Make use of the parabola of safety.] A mortar gun is placed on the summit of a conical hill of height \(60 \mathrm{~m}\) and base diameter \(160 \mathrm{~m}\). If the gun has a muzzle speed of \(25 \mathrm{~m} \mathrm{~s}^{-1}\), can it shell anywhere on the hill? [Take \(\left.g=10 \mathrm{~m} \mathrm{~s}^{-2} .\right]\)

A mortar gun, with a maximum range of \(40 \mathrm{~m}\) on level ground, is placed on the edge of a vertical cliff of height \(20 \mathrm{~m}\) overlooking a horizontal plain. Show that the horizontal range \(R\) of the mortar gun is given by $$R=40\left\\{\sin \alpha+\left(1+\sin ^{2} \alpha\right)^{\frac{1}{2}}\right\\} \cos \alpha$$ where \(\alpha\) is the angle of elevation of the mortar above the horizontal. [Take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\).] Evaluate \(R\) (to the nearest metre) when \(\alpha=45^{\circ}\) and \(35^{\circ}\) and confirm that \(\alpha=45^{\circ}\) does not yield the maximum range. [Do not try to find the optimum projection angle this way. See Problem \(4.22\) below. \(]\)

A body of mass \(m\) is projected with speed \(u\) in a medium that exerts a resistance force of magnitude (i) \(m k|v|\), or (ii) \(m K|v|^{2}\), where \(k\) and \(K\) are positive constants and \(v\) is the velocity of the body. Gravity can be ignored. Determine the subsequent motion in each case. Verify that the motion is bounded in case (i), but not in case (ii).

A particle is projected from the origin with speed \(u\) in a direction making an angle \(\alpha\) with the horizontal. The motion takes place in the \((x, z)\)-plane, where \(O z\) points vertically upwards. If the projection speed \(u\) is fixed, show that the particle can be made to pass through the point \((a, b)\) for some choice of \(\alpha\) if \((a, b)\) lies below the parabola $$ z=\frac{u^{2}}{2 g}\left(1-\frac{g^{2} x^{2}}{u^{4}}\right) $$ This is called the parabola of safety. Points above the parabola are 'safe' from the projectile. An artillery shell explodes on the ground throwing shrapnel in all directions with speeds of up to \(30 \mathrm{~m} \mathrm{~s}^{-1}\). A man is standing at an open window \(20 \mathrm{~m}\) above the ground in a building \(60 \mathrm{~m}\) from the blast. Is he safe? [Take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\).]

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