Problem 12 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

If we ignore air resistance, a baseball thrown from shoulder level at an angle of \(\theta\) radians with the ground and at an initial velocity of \(v_{0}\) meters per second will be at shoulder level again when it is \(\frac{v_{0}^{2} \sin (2 \theta)}{g}\) meters away. \(g\) is the acceleration due to gravity \((9.8\) \(\left.\mathrm{m} / \mathrm{sec}^{2}\right)\) (a) Express the maximum distance the baseball can travel (from shoulder level to shoulder level) in terms of the initial velocity. (b) The fastest baseball pitchers can throw about 100 miles per hour. How far would such a ball travel if thrown at the optimal angle? (Note: 1 mile \(=5280\) feet and 1 meter \(\approx 3.28\) feet. \()(*)\)

Step-by-Step Solution

Answer

a) The maximum distance the baseball can travel from shoulder level to shoulder level is \(\frac{v_{0}^{2}}{g}\) meters. b) The baseball, thrown at a speed of 100 miles per hour, can travel approximately 203 meters.

Step 1: Express the maximum distance

The maximum range of a projectile occurs when \(\theta\) = \(\pi / 4\). Substituting this into the given formula, we get the maximum distance \(d_{max}\) as \(\frac{v_{0}^{2} \sin (2 \cdot \pi / 4)}{g}\) = \(\frac{v_{0}^{2}}{g}\)

Step 2: Convert units

First, convert the speed from miles per hour to feet per second using the conversion 1 mph = 1.467 feet per second. Then convert feet per second to meters per second using the conversion 1 foot = 0.3048 meters. So, 100 miles per hour is 100 x 1.467 x 0.3048 = 44.704 meters per second

Step 3: Substitute velocity into the formula

Substitute this velocity into the formula from step 1 to get the maximum distance a baseball can travel when thrown at the optimal angle: \(d_{max}\) = \(\frac{44.704^{2}}{9.8}\) meters

Key Concepts

Optimal Angle for Maximum Range

Understanding the optimal angle for maximum range in projectile motion is crucial in many fields including sports, engineering, and military applications. When launching a projectile, such as throwing a baseball or firing a cannonball, the distance it travels horizontally before hitting the ground is its range.

Through calculus, we can derive that the angle which maximizes this range is 45 degrees, or \( \frac{\pi}{4} \) radians. This is because the range equation, given as \( d = \frac{v_{0}^{2} \sin (2 \theta)}{g} \) where \( \theta \) is the angle of launch, \( v_{0} \) is the initial velocity, and \( g \) is the acceleration due to gravity, reaches its maximum when \( \sin(2 \theta) \) is at its maximum value, which is 1.

When \( \theta = \frac{\pi}{4} \) radians, \( 2 \theta \) equals \( \frac{\pi}{2} \) radians and that corresponds to the optimal launch angle because \( \sin \frac{\pi}{2} = 1 \)—yielding the maximum possible range. In the context of our baseball exercise, we used this knowledge to deduce that the maximum distance can be achieved when throwing the ball at this precise angle.

Converting Units in Physics

Converting units in physics is a fundamental skill that ensures the validity of our calculations across different measurement systems. Physics problems, like our projectile motion example, may present values in a mix of units that need to be standardized to perform accurate calculations.

Why is Unit Conversion Important?

Unit conversion is critical because equations in physics require consistency in the units used. For instance, if the acceleration due to gravity is given in meters per second squared (\(m/s^2\)), and the initial velocity is given in miles per hour (mph), we cannot directly substitute these into our equations without converting them to compatible units. Discrepancies in units can lead to incorrect results and a misunderstanding of the physical concept.

The step by step solution converts the pitcher's throw from miles per hour to meters per second, which is necessary for use with the acceleration due to gravity (\(g\)) unit. This process often involves multiple conversion factors, as seen with the conversion from mph to feet per second, and then to meters per second. It's a two-step process: we multiply by the conversion factor from mph to fps, then multiply by the conversion factor from feet to meters.

Acceleration due to Gravity

The acceleration due to gravity, represented by the symbol \( g \), is the constant acceleration experienced by an object in free fall near the Earth's surface. This value is approximately \( 9.8 \ m/s^2 \). Important for understanding projectile motion, \( g \) dictates the vertical motion's acceleration, profoundly influencing the object's trajectory.

The Role of Gravity in Projectile Motion

In the absence of air resistance, gravity is the only force acting on a projectile in the vertical direction. Hence, it affects the time the projectile spends in the air and its vertical displacement. When calculating the range, as in our baseball example, \( g \) appears in the denominator. This demonstrates that a higher acceleration due to gravity would result in a shorter range for the same initial velocity. Understanding how \( g \) affects projectile motion can apply to various practical scenarios such as golf ball trajectories, or even when calculating the orbits of satellites in space.