Problem 20 - Algebra 2

Question

For Exercises 20 and \(21,\) use the following information. If an object is propelled from ground level, the maximum height that it reaches is given by \(h=\frac{v^{2} \sin ^{2} \theta}{2 g},\) where \(\theta\) is the angle between the ground and the initial path of the object, \(v\) is the object's initial velocity, and \(g\) is the acceleration due to gravity, 9.8 meters per second squared. Verify the identity \(\frac{v^{2} \sin ^{2} \theta}{2 g}=\frac{v^{2} \tan ^{2} \theta}{2 g \sec ^{2} \theta}\).

Step-by-Step Solution

Answer

The identity is verified; both expressions are equal.

Step 1: Understand the Given Identity

We are given the equation \(h = \frac{v^2 \sin^2 \theta}{2g}\) and we need to verify if it is equivalent to \(\frac{v^2 \tan^2 \theta}{2g \sec^2 \theta}\). Our goal is to show that both expressions for height \(h\) are equivalent.

Step 2: Simplify the Right-Hand Side

The right-hand side of the identity is \(\frac{v^2 \tan^2 \theta}{2g \sec^2 \theta}\). Recall that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and \(\sec \theta = \frac{1}{\cos \theta}\). Substitute these into the expression:\[\frac{v^2 \left(\frac{\sin \theta}{\cos \theta}\right)^2}{2g \left(\frac{1}{\cos \theta}\right)^2} = \frac{v^2 \frac{\sin^2 \theta}{\cos^2 \theta}}{2g \frac{1}{\cos^2 \theta }}.\]

Step 3: Simplify Further

The \(\cos^2 \theta\) in the denominator of the division cancels out:\[\frac{v^2 \frac{\sin^2 \theta}{\cos^2 \theta}}{2g \frac{1}{\cos^2 \theta}} = \frac{v^2 \sin^2 \theta \cdot \cos^2 \theta}{2g \cdot \cos^2 \theta} = \frac{v^2 \sin^2 \theta}{2g}.\]

Step 4: Conclude the Verification

Through simplification, we have shown:\[\frac{v^2 \tan^2 \theta}{2g \sec^2 \theta} = \frac{v^2 \sin^2 \theta}{2g}.\]Both sides of the identity are equal, thus verifying the identity given in the problem.

Key Concepts

maximum height formula

The maximum height formula is a key concept when understanding projectile motion and it plays a crucial role in calculating how high an object can go when thrown or propelled from the ground. Imagine you are launching a ball into the air. The higher the ball goes, the more intriguing the physics behind it looks. The formula to find this maximum height is given by:

• \( h = \frac{v^{2} \sin ^{2} \theta}{2g} \)

where:

• \( h \) is the maximum height,
• \( v \) is the initial velocity of the object,
• \( \sin \theta \) represents the sine of the angle \( \theta \),
• \( g \) is the acceleration due to gravity which is \( 9.8 \) meters per second squared.

This formula shows that the maximum height depends on the square of the velocity and the sine of the angle squared. Essentially, a higher velocity or a steeper angle will allow an object to reach greater heights. It's remarkable how a simple equation can tell so much about the motion of an object. You can use this formula in various practical situations, such as sports, engineering, and even launching rockets.

projectile motion

Projectile motion is a fascinating topic in physics that describes the motion of an object that is launched into the air and subject to gravity. When you throw a ball, it follows a curved path due to these forces. This type of motion can be broken down into two components:

• Horizontal motion, which is uniform and constant.
• Vertical motion, which is affected by gravity.

Together, these create a projectile path, or trajectory, that is symmetrical and parabolic. By studying projectile motion, you can predict where an object will land, how high it will travel, and how long it will stay in the air.
The study of projectile motion finds applications in various fields, such as sports where you might calculate the best angle to kick a football, or in engineering where you need to know how an object will travel before it impacts. Being able to determine the trajectory and maximum height is critical, and it often involves using trigonometric identities and equations, like our maximum height formula.

angle of projection

The angle of projection is a vital component in determining the behavior of a projectile. It is the angle \( \theta \) at which an object is launched concerning the ground. Understanding this angle allows one to predict several aspects of the object's motion. This angle affects:

• The range of the projectile, which is how far it travels horizontally.
• The maximum height, as a larger angle can lead to a higher arc.
• The time of flight, indicating how long the projectile will be in motion.

The ideal angle of projection for maximum range without air resistance is typically \( 45^{\circ} \). Yet, if you aim for maximum height, a steeper angle is often more advantageous. The element of trigonometry in these calculations is evident through expressions like \( \sin \theta \) and \( \tan \theta \), which illustrate how adjusting the angle can manipulate the outcomes of a projectile's journey. Understanding the angle of projection is crucial for anyone engaging in activities or studies that involve launching objects, from sports to spacecraft.