Problem 57 - Calculus Single Variable

Question

An arrow is shot at a \(60^{\circ}\) angle to the horizontal with initial velocity \(190 \mathrm{ft} / \mathrm{s}\). How high will the arrow travel? What will be the horizontal component of its velocity at height 50 feet (going up)?

Step-by-Step Solution

Answer

The maximum height is approximately 422.70 feet. The horizontal velocity at 50 feet is 95 ft/s.

Step 1: Identify and Decompose the Initial Velocity

First, decompose the initial velocity of the arrow into its horizontal and vertical components. Given the angle \( \theta = 60^{\circ} \) and the initial velocity \( v_0 = 190 \; \text{ft/s} \), the vertical component, \( v_{0y} \), is \( v_{0y} = v_0 \sin(\theta) \), and the horizontal component, \( v_{0x} \), is \( v_{0x} = v_0 \cos(\theta) \).

Step 2: Calculate the Vertical Component

Using the formula from Step 1, compute the vertical component: \( v_{0y} = 190 \sin(60^\circ) = 190 \times \frac{\sqrt{3}}{2} \approx 164.55 \; \text{ft/s} \).

Step 3: Determine the Maximum Height

To find the maximum height, use the formula \( v_y^2 = v_{0y}^2 - 2g h \), where \( v_y = 0 \) at maximum height, \( g = 32 \; \text{ft/s}^2 \) is the acceleration due to gravity, and \( h \) is maximum height. Solving for \( h \), we get \[ h = \frac{v_{0y}^2}{2g} = \frac{(164.55)^2}{2 \times 32} \approx 422.70 \; \text{feet} \].

Step 4: Calculate the Horizontal Component at 50 Feet

At 50 feet, the vertical component \( v_y \) is found using \( v_y^2 = v_{0y}^2 - 2g h' \), where \( h' = 50 \). Calculate \( v_y \) as \( v_y = \sqrt{164.55^2 - 2 \times 32 \times 50} \approx 152.50 \; \text{ft/s} \). Since gravity does not affect the horizontal velocity, \( v_x = v_{0x} = 190 \cos(60^\circ) = 190 \times \frac{1}{2} = 95 \; \text{ft/s} \).

Key Concepts

Kinematics

Kinematics is the branch of classical mechanics that describes the motion of points, objects, and systems of objects without considering the causes of motion (such as forces). When analyzing projectile motion, such as an arrow being shot, it is crucial to understand the basic principles of kinematics. This motion involves the arrow moving through two dimensions: horizontal and vertical. In this context, we are often interested in quantities like velocity, displacement, and acceleration.

• Velocity: The speed of the arrow in a particular direction.
• Acceleration: This problem only considers gravitational acceleration acting downward.
• Displacement: How far the arrow travels in both horizontal and vertical directions.

Grasping the foundational concepts of kinematics helps us analyze how the projectile's motion evolves over time, understanding both its peak height and reach.

Velocity Decomposition

To study an object's projectile motion in detail, it is essential to decompose its velocity into horizontal and vertical components. This process, known as velocity decomposition, is pivotal in solving projectile problems. When an arrow is shot at an angle, we split the initial velocity into two parts that act at right angles to each other:

• Horizontal Component (\(v_{0x}\)): Determined using \(v_0 \cos(\theta)\), this represents the constant velocity in the horizontal direction.
• Vertical Component (\(v_{0y}\)): Calculated with \(v_0 \sin(\theta)\), this reflects the initial upward velocity affected by gravity.

For our arrow, the initial velocity is 190 ft/s at a 60-degree angle. By applying basic trigonometry, we can find:

• \(v_{0x} = 190 \cos(60^\circ) = 95 \; \text{ft/s}\)
• \(v_{0y} = 190 \sin(60^\circ) = 164.55 \; \text{ft/s}\)

These components allow us to analyze the projectile more effectively.

Trigonometry

Trigonometry is the mathematical study of the relationships between the angles and sides of triangles. In projectile motion, trigonometry provides us with the means to decompose velocities based on angles. This concept allows us to find the horizontal and vertical components of an initial velocity vector. Let's break it down:

• Sine Function: Relates the angle to the opposite side over the hypotenuse in a right triangle, utilized as \(\sin(\theta)\) for computing vertical components.
• Cosine Function: Relates the angle to the adjacent side over the hypotenuse, used as \(\cos(\theta)\) for determining horizontal components.

In the problem at hand, knowing \(\theta = 60^\circ\) and \(v_0 = 190 \mathit{ft/s}\) connects us to these trigonometric equations:

• Vertical: \(v_{0y} = 190 \sin(60^\circ)\)
• Horizontal: \(v_{0x} = 190 \cos(60^\circ)\)

This application of trigonometric principles is critical for unraveling projectile dynamics.

Maximum Height

Determining the maximum height of a projectile is a key aspect of studying its vertical motion. At the maximum height, the vertical velocity of the projectile is zero because it has momentarily stopped ascending and is about to descend. To find this, we leverage the kinematic equation:\[ h = \frac{v_{0y}^2}{2g} \]where \(v_{0y}\) is the initial vertical velocity, and \(g\) is the acceleration due to gravity (32 ft/s²).
From our calculations, with \(v_{0y} = 164.55 \mathit{ft/s}\):\[ h = \frac{(164.55)^2}{2 \times 32} \approx 422.70 \text{ feet} \]This result shows the maximum height reached by the arrow. Understanding this concept helps in determining how high a projectile will travel before gravity pulls it back down.