Problem 41 - Calculus

Question

A baseball hit at an angle of \(\theta\) to the horizontal with initial velocity \(v_{0}\) has horizontal range, \(R,\) given by $$R=\frac{v_{0}^{2}}{g} \sin (2 \theta)$$ Here \(g\) is the acceleration due to gravity. Sketch \(R\) as a function of \(\theta\) for \(0 \leq \theta \leq \pi / 2 .\) What angle gives the maximum range? What is the maximum range?

Step-by-Step Solution

Answer

The maximum range is at 45 degrees with \( R_{max} = \frac{v_{0}^{2}}{g} \).

Step 1: Understand the formula

The horizontal range \( R \) of a projectile is given by \( R = \frac{v_{0}^{2}}{g} \sin(2\theta) \), where \( v_0 \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( \theta \) is the launch angle. This formula shows that \( R \) is directly dependent on the value of \( \sin(2\theta) \).

Step 2: Graph the function

Since \( R = \frac{v_{0}^{2}}{g} \sin(2\theta) \), to graph \( R \) as a function of \( \theta \), we first consider the behavior of \( \sin(2\theta) \) over the interval \( 0 \leq \theta \leq \pi/2 \). The function \( \sin(2\theta) \) reaches its maximum value of 1 when \( 2\theta = \frac{\pi}{2} \), or \( \theta = \frac{\pi}{4} \). Therefore, \( R \) will have its maximum value at this angle.

Step 3: Determine the maximum angle

From the graph and the behavior of the sine function, the maximum value is at \( \theta = \frac{\pi}{4} \) or 45 degrees, which gives the maximum range. This is because \( \sin(2\theta) = 1 \) at this angle.

Step 4: Calculate the maximum range

Substitute \( \theta = \frac{\pi}{4} \) into the range formula to find the maximum range: \[ R_{max} = \frac{v_{0}^{2}}{g} \sin(2(\frac{\pi}{4})) = \frac{v_{0}^{2}}{g} \times 1 = \frac{v_{0}^{2}}{g} \] So, the maximum range \( R_{max} \) is achieved at an angle of 45 degrees.

Key Concepts

Trigonometric Functions

Trigonometric functions play an essential role in understanding projectile motion, especially when dealing with launch angles and maximum ranges. In the given problem, the sine function, specifically \( \sin(2\theta) \), is crucial. This trigonometric function describes the relationship between the angle at which a projectile is launched and the resulting range it will cover.

The trigonometric function \( \sin(2\theta) \) oscillates between -1 and 1. For our purposes, since we're dealing with angles between 0 and \( \pi/2 \) radians (or 0 to 90 degrees), the sine function only takes on non-negative values.

Key points to remember include:

• \( \sin(0) = 0 \)
• \( \sin(\pi/2) = 1 \)
• \( \sin(\pi) = 0 \)

These values will help you predict how the range \( R \) will behave as the angle \( \theta \) changes, especially since the range directly relies on \( \sin(2\theta) \)'s variability.

Understanding how to graph this trigonometric function from \( 0 \) to \( \pi/2 \) is fundamental in determining when and how often the range is maximized.

Maximum Range

To achieve maximum range in projectile motion, the role of the initial velocity \( v_0 \) and the acceleration due to gravity \( g \) must be clearly understood, but most importantly, we focus on how the sine function governs this.

Maximum range occurs when the expression \( \sin(2\theta) \) reaches its peak value of 1. Mathematically, this is when \( 2\theta = \pi/2 \), translating to \( \theta = \pi/4 \) or 45 degrees. At this angle, the range \( R \) is at its highest potential, given by the formula:\[ R_{max} = \frac{v_{0}^{2}}{g} \]
This simple yet powerful equation shows that achieving the maximum range depends not just on choosing the right angle, but also on having sufficient initial velocity to overcome gravitational pull.

In practical applications, such as sports or ballistics, understanding these conditions enables better performance and accuracy.

Launch Angle

The launch angle \( \theta \) is a pivotal factor in projectile motion that directly influences the path and distance a projectile travels. It specifies direction relative to the horizontal plane when the projectile is launched.

Why is the launch angle important? Because changing \( \theta \) alters how the initial velocity \( v_0 \) is divided into horizontal and vertical components, thus influencing the distance traveled. At \( \theta = 45^\circ \), the horizontal and vertical components are balanced, optimizing the range.

The effects of varying launch angles include:

• A low launch angle results in more horizontal distance but less vertical height.
• A high launch angle increases vertical height but can decrease horizontal distance.

Balancing these factors by selecting the optimal launch angle, typically 45 degrees for maximum range, provides the ideal solution in projectile motion scenarios.

This understanding is crucial whether you're solving physics problems or applying the concepts to real-world tasks like setting up a golf shot or firing a cannon.