Problem 83 - Precalculus with Limits

Question

Velocity A gun with a muzzle velocity of 1200 feet per second is fired at an angle of \(6^{\circ}\) above the horizontal. Find the vertical and horizontal components of the velocity.

Step-by-Step Solution

Answer

The horizontal component of the velocity is approximately \(1196.5\) ft/sec, and the vertical component is approximately \(125.6\) ft/sec.

Step 1: Find the horizontal component of velocity

The horizontal, or x-component, of the velocity can be calculated by multiplying the original velocity by the cosine of the shot angle: \(v_x = v \cdot \cos(\theta)\) where \(v\) is the original velocity (1200 feet per second) and \(\theta\) is the angle of elevation (6 degrees).

Step 2: Calculate the numerical horizontal velocity

Insert the given values into the equation to calculate the horizontal velocity: \(v_x = 1200 \cdot \cos(6^{\circ})\). After calculating, approximately, we have \(v_x = 1196.5\) ft/sec.

Step 3: Find the vertical component of velocity

The vertical, or y-component, of the velocity can be calculated by multiplying the original velocity by the sine of the shot angle: \(v_y = v \cdot \sin(\theta)\) where \(v\) is the original velocity (1200 feet per second) and \(\theta\) is the angle of elevation (6 degrees).

Step 4: Calculate the numerical vertical velocity

Insert the given values into the equation to calculate the vertical velocity: \(v_y = 1200 \cdot \sin(6^{\circ})\). After calculating, approximately, we have \(v_y = 125.6\) ft/sec.

Key Concepts

Horizontal Velocity Component

In projectile motion, the horizontal velocity component is crucial because it describes how fast an object moves along the horizontal axis. When a projectile is fired, like the gun mentioned in the exercise, its initial velocity can be broken into two parts: horizontal and vertical. The horizontal component is determined by the angle at which the projectile is launched and the overall speed of the projectile.

To find the horizontal velocity, we use the trigonometric function cosine. This involves multiplying the initial velocity by the cosine of the angle of elevation. In mathematical terms, it's represented as:

• \[ v_x = v \cdot \cos(\theta) \]

where:

• \( v_x \) is the horizontal velocity component
• \( v \) is the total initial velocity (1200 ft/sec in this case)
• \( \theta \) is the launch angle (6° for this scenario)

Using this formula ensures that we can accurately determine how fast the projectile travels along the horizontal path, which is approximately 1196.5 ft/sec in this instance.

Vertical Velocity Component

While the horizontal velocity component affects how far a projectile travels, the vertical velocity component determines how high it will go. To calculate the vertical velocity, you use the sine function from trigonometry. This is because the vertical component relates to the opposite side of the angle when visualizing the projectile's motion as part of a right triangle.

To compute this component, you apply the following formula:

• \[ v_y = v \cdot \sin(\theta) \]

where:

• \( v_y \) is the vertical velocity component
• \( v \) remains the initial velocity
• \( \theta \) is the angle of elevation

For our given example, once the values are plugged into the equation, you get a vertical velocity of approximately 125.6 ft/sec. Understanding the vertical velocity helps in predicting how the projectile will move upwards and then fall back down under the influence of gravity.

Trigonometric Functions

Trigonometric functions, namely sine and cosine, are integral to decomposing motion into horizontal and vertical components in projectile motion problems. These functions are based on the properties of right-angled triangles and are essential because they allow us to handle angles succinctly and precisely.

Sine (

• \( \sin \), represents the ratio of the length of the opposite side to the hypotenuse in a right triangle. It helps calculate vertical components by reflecting how much of the initial velocity is directed vertically.

Cosine (

• \( \cos \), depicts the ratio of the length of the adjacent side to the hypotenuse. It quantifies the horizontal component by showing how much velocity is parallel to the ground level.

The importance of these functions in projectile motion cannot be overstated; without them, it would be challenging to break down the motion into comprehensible parts that can be individually analyzed. They help to understand and predict how fast, high, and far an object can travel when projected at an angle.