Problem 46 - Thomas Calculus

Question

Projectile motion The range \(R\) of a projectile fired from the origin over horizontal ground is the distance from the origin to the point of impact. If the projectile is fired with an initial velocity \(v_{0}\) at an angle \(\alpha\) with the horizontal, then in Chapter 13 we find that \begin{equation}R=\frac{v_{0}^{2}}{g} \sin 2 \alpha,\end{equation} where \(g\) is the downward acceleration due to gravity. Find the angle \(\alpha\) for which the range \(R\) is the largest possible.

Step-by-Step Solution

Answer

The angle \(\alpha\) for maximum range is \(45^\circ\) or \(\frac{\pi}{4}\).

Step 1: Understanding the Problem

We're asked to find the angle \(\alpha\) that maximizes the range \(R\) of a projectile, given that \( R = \frac{v_{0}^{2}}{g} \sin 2\alpha \). The goal is to find \(\alpha\) that makes \(R\) as large as possible.

Step 2: Identify the Function to Maximize

We need to maximize \(R = \frac{v_{0}^{2}}{g} \sin 2\alpha\). Since \(\frac{v_{0}^{2}}{g}\) is a constant, we focus on maximizing \(\sin 2\alpha\).

Step 3: Determine Maximum of Sine Function

The function \(\sin 2\alpha\) reaches its maximum value of 1 when \(2\alpha = \frac{\pi}{2} + 2k\pi\), where \(k\) is any integer. In simple terms, the maximum value of \(\sin 2\alpha\) is 1.

Step 4: Solve for \(\alpha\)

From \(2\alpha = \frac{\pi}{2}\), we solve for \(\alpha\) by dividing both sides by 2, giving \(\alpha = \frac{\pi}{4}\).

Key Concepts

Range of a Projectile

The range of a projectile is a fascinating concept in physics. It represents the horizontal distance that a projectile travels before it hits the ground. To calculate this distance, we use a formula that connects several key variables of the projectile's motion. One of the main variables involved is the initial velocity, denoted as \(v_0\). Another is the angle of projection, \(\alpha\), which is the angle at which the projectile is launched. The formula also includes \(g\), the acceleration due to gravity, typically valued at about \(9.8\, \text{m/s}^2\) on Earth. This constant pulls the projectile down to the ground.

The mathematical expression for the range \(R\) is:

• \(R = \frac{v_{0}^{2}}{g} \sin 2\alpha\)

This equation tells us that the range is directly proportional to the square of the initial velocity and depends on the sine of twice the angle of projection \(\alpha\). Understanding this relationship is key when aiming for the maximum range. The sine factor shows that there's a cyclic nature to how the angle affects the range. This is where our next topic comes into play: optimizing this parameter.

Angle of Projection

The angle of projection, \(\alpha\), plays a crucial role in determining how far a projectile will travel. It is the angle between the initial velocity vector and the horizontal axis. When adjusting this angle, you are essentially changing how you balance the projectile's vertical and horizontal velocity components.

Manipulating \(\alpha\) affects the projectile's trajectory and, consequently, the range. In this context, only certain angles will maximize the distance a projectile travels. Through the formula \(R = \frac{v_{0}^{2}}{g} \sin 2\alpha\), we can see that the sine function, \(\sin 2\alpha\), influences the range.

By analyzing \(\sin 2\alpha\), we can find the optimal angle for maximum distance. The sine function reaches its highest value of \(1\). So, for \(\sin 2\alpha\), the maximum occurs when \(2\alpha = \frac{\pi}{2}\), which simplifies to \(\alpha = \frac{\pi}{4}\) or 45 degrees. This means launching a projectile at 45 degrees provides the peak range on level ground at initial launch conditions.

Trigonometric Optimization

When it comes to optimizing the range, trigonometry becomes an essential tool. We've seen that the range is influenced by \(\sin 2\alpha\), a function with well-known properties.

For optimization, understanding the maximum value of sine functions is critical. The sine of an angle reaches a maximum of \(1\) at specific points in its cycle. For \(\sin x\), that point is when \(x = \frac{\pi}{2} + 2k\pi\), where \(k\) is any integer.

This property of sine enables us to determine the optimal angle \(\alpha\), maximizing the range \(R\). We solve \(2\alpha = \frac{\pi}{2}\) to find \(\alpha\). Consequently, this results in \(\alpha = \frac{\pi}{4}\), an angle commonly known as 45 degrees when converted to degrees.

Thus, mastering trigonometric principles provides powerful insights and solutions to maximize the effects of projectile motion.