Question

The horizontal distance, \(d,\) in metres, travelled by a ball that is kicked at an angle, \(\theta,\) with the ground is modelled by the formula \(d=\frac{2\left(v_{0}\right)^{2} \sin \theta \cos \theta}{g},\) where \(V_{0}\) is the initial velocity of the ball, in metres per second, and \(g\) is the force of gravity \(\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)\) a) Rewrite the formula using a double-angle identity. b) Determine the angle \(\theta \in\left(0^{\circ}, 90^{\circ}\right)\) that would result in a maximum distance for an initial velocity \(v_{0}\). c) Explain why it might be easier to answer part b) with the double-angle version of the formula that you determined in part a).

Short answer

a) \( d = \frac{\left(v_{0}\right)^{2} \sin 2\theta}{g} \). b) \( \theta = 45^{\circ} \). c) Simplifies finding the maximum distance due to a single term.

Step-by-step solution

Rewrite using a double-angle identity

The given formula is \[ d = \frac{2 \left(v_{0}\right)^{2} \sin \theta \cos \theta}{g} \]Use the double-angle identity for sine: \[ \sin 2\theta = 2 \sin \theta \cos \theta \]Replace in the formula: \[ d = \frac{\left(v_{0}\right)^{2} \sin 2\theta}{g} \].

Determine angle for maximum distance

The distance formula in double-angle form is \[ d = \frac{\left(v_{0}\right)^{2} \sin 2\theta}{g} \]To maximize the distance, we need to maximize \( \sin 2\theta \). The maximum value of \( \sin x \) is 1. So, \( \sin 2\theta = 1 \).This occurs when \( 2\theta = 90^{\circ} \) or \( \theta = 45^{\circ} \).

Explanation

Using the formula \[ d = \frac{2 \left(v_{0}\right)^{2} \sin \theta \cos \theta}{g} \]directly involves both sine and cosine terms, making it harder to identify the maximum distance conditions.By converting it to \[ d = \frac{\left(v_{0}\right)^{2} \sin 2\theta}{g} \], we simplify the problem to finding the maximum of the single term \( \sin 2\theta \), which is straightforward.

Key concepts

Double-Angle Identity

In trigonometry, double-angle identities help us simplify expressions involving trigonometric functions. When dealing with projectile motion, these identities can make complex formulas easier to work with. For example, the given formula for the horizontal distance traveled by a ball is:

\rightarrow \[ d = \frac{2 \left(v_{0}\right)^{2} \sin \theta \cos \theta}{g} \] \rightarrow By applying the double-angle identity for sine, which is: \[ \sin 2\theta = 2 \sin \theta \cos \theta \],
we can simplify the formula to: \[ d = \frac{\left(v_{0}\right)^{2} \sin 2\theta}{g} \] This simplification is crucial because it enables us to focus on a single trigonometric function instead of managing two terms (sine and cosine). When solving for the maximum distance, working with just one term, \( \sin 2\theta \), becomes much easier. Remember, recognizing and using double-angle identities can save you a lot of time and effort in trigonometry problems.

Maximizing Sine Function

To find the angle that maximizes the horizontal distance, we need to maximize the sine function involved in our formula. When we rewrite the formula using the double-angle identity, it turns into:

\rightarrow \[ d = \frac{\left(v_{0}\right)^{2} \sin 2\theta}{g} \] Here, the distance \(d\) is directly proportional to \( \sin 2\theta \). The maximum value of the sine function is 1. Therefore, to achieve maximum distance, we need: \( \sin 2\theta = 1 \).
This is true when: \( 2\theta = 90^{\circ} \), which implies \( \theta = 45^{\circ} \).

In the context of projectile motion, this means that the angle of 45 degrees with the ground is ideal for maximizing how far the ball will travel. Insights like these are what make understanding the behavior of trigonometric functions so powerful in physics and engineering problems.

Trigonometric Formulas

Trigonometric formulas are essential tools in both mathematics and physics. Let's break down how they help solve real-world problems, like calculating distances in projectile motion. In this problem, you're dealing with the formula for horizontal distance: \[ d = \frac{2 \left(v_{0}\right)^{2} \sin \theta \cos \theta}{g} \].
This formula combines the functions \( \sin \theta \) and \( \cos \theta \). Using the double-angle identity, it simplifies to: \[ d = \frac{\left(v_{0}\right)^{2} \sin 2\theta}{g} \].
Here’s a quick recap of key trigonometric identities that are helpful:

• Double-Angle Identity: \(\sin 2\theta = 2 \sin \theta \cos \theta\)
• Basic Sine and Cosine Values: \(\sin 90^{\circ} = 1\), \(\cos 0^{\circ} = 1\).

These identities are used to convert tricky expressions into simpler ones, making calculations easier. Remember that understanding these formulas not only helps you solve textbook problems but also equips you with the skills to tackle more complex scenarios involving angles and distances.