Problem 57 - Precalculus

Question

Bullet Speed. A bullet is fired from ground level at a speed of 2200 feet per second at an angle of \(30^{\circ}\) from the horizontal. Find the magnitude of the horizontal and vertical components of the velocity vector.

Step-by-Step Solution

Answer

Horizontal velocity is 1905.2 ft/s, vertical velocity is 1100 ft/s.

Step 1: Identify the Components of Velocity

When a bullet is fired at an angle, its velocity can be broken down into two perpendicular components: horizontal (x-direction) and vertical (y-direction). We will use trigonometric functions to determine these components.

Step 2: Calculate the Horizontal Component of Velocity

The horizontal component of velocity, denoted as \(v_x\), is found using the cosine of the angle. It is calculated using the formula \(v_x = v \cdot \cos \theta\), where \(v\) is the initial velocity and \(\theta\) is the angle of projection. For this exercise:\[ v_x = 2200 \cdot \cos(30^{\circ}) \]

Step 3: Calculate \(\cos(30^{\circ})\)

Using the known trigonometric identity, \(\cos(30^{\circ}) = \frac{\sqrt{3}}{2}\), we substitute into the formula for the horizontal component:\[ v_x = 2200 \cdot \frac{\sqrt{3}}{2} \]

Step 4: Solve for Horizontal Component

Compute the horizontal component:\[ v_x = 2200 \cdot 0.866 = 1905.2 \text{ feet per second} \]

Step 5: Calculate the Vertical Component of Velocity

The vertical component of velocity, denoted as \(v_y\), is found using the sine of the angle. It is calculated using the formula \(v_y = v \cdot \sin \theta\).For this exercise:\[ v_y = 2200 \cdot \sin(30^{\circ}) \]

Step 6: Calculate \(\sin(30^{\circ})\)

Using the known trigonometric identity, \(\sin(30^{\circ}) = \frac{1}{2}\), we substitute into the formula for the vertical component:\[ v_y = 2200 \cdot \frac{1}{2} \]

Step 7: Solve for Vertical Component

Compute the vertical component:\[ v_y = 2200 \cdot 0.5 = 1100 \text{ feet per second} \]

Key Concepts

Trigonometric Functions

Imagine you have a tool that helps you break angles and directions into smaller, digestible parts. Trigonometric functions are exactly that tool! When a projectile like a bullet is thrown at an angle, it can be difficult to measure its movement directly.
To make it simpler, we use **trigonometric functions**, such as *sine* and *cosine*, to split the initial velocity into horizontal and vertical components. This involves applying simple mathematical formulas.

• **Cosine**: Helps us find the horizontal component by measuring how much of the velocity is moving along the ground.
• **Sine**: Determines the vertical component by showing how high up the velocity is moving.

Thus, trigonometric functions transform complicated projectile paths into two simpler perpendicular paths.

Velocity Components

Understanding velocity components is key to breaking down projectile motion clearly. The initial velocity of a projectile, like a bullet, can be split into two components: horizontal and vertical.

• **Horizontal Velocity Component**: Often denoted as \(v_x\), this component remains constant if we ignore air resistance. It's calculated using the formula \(v_x = v \cdot \cos \theta\), where \(v\) is the initial speed and \(\theta\) is the angle of projection.
• **Vertical Velocity Component**: Denoted as \(v_y\), this component is subject to changes because it’s affected by gravity. The formula to find this component is \(v_y = v \cdot \sin \theta\).

By calculating these components, you get a clear picture of how the bullet moves both horizontally along the ground and vertically into the air. This decomposition is critical for predicting the projectile's path.

Angle of Projection

Think of the angle of projection as the direction in which you aim a slingshot. The projectile's path depends heavily on this angle. The angle of projection not only decides the initial height of the projectile but also affects how far it will go.
An angle of \(30^{\circ}\), as in our bullet example, typically results in a medium trajectory, where neither height nor distance is maximized but balanced.

• **Understanding the Importance**: Different angles will split a projectile's velocity differently, affecting both its range and maximum height.
• **Practical Implications**: Knowing how to adjust this angle is crucial in applications like ballistics, sports, and engineering.

Choosing the right angle can optimize how far or high a projectile travels, making this a fundamental concept in physics and engineering.