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Chapter 7: Problem 83

An arrow is shot into the air at an angle of \(37^{\circ}\) with an initial velocity of \(100 \mathrm{ft} / \mathrm{sec}\). Compute the horizontal and vertical components of the representative vector.

Short Answer

Expert verified
Horizontal: 79.86 ft/sec, Vertical: 60.18 ft/sec.

Step by step solution

01

Identify the Given Information

We are given that the arrow is shot at an initial velocity of \( v = 100 \) ft/sec and at an angle of \( \theta = 37^{\circ} \) from the horizontal. From this, we aim to find the horizontal and vertical components of the velocity.
02

Understand the Component Formulas

The formulas to resolve a vector into its components are: \[v_x = v \cdot \cos(\theta)\quad \text{and}\quad v_y = v \cdot \sin(\theta)\] where \( v_x \) is the horizontal component and \( v_y \) is the vertical component.
03

Calculate the Horizontal Component

Substitute the given values into the formula for the horizontal component: \[v_x = 100 \cdot \cos(37^{\circ})\] We need to use the cosine of \(37^{\circ}\), which is approximately \(0.7986\). Thus, \[v_x = 100 \cdot 0.7986 = 79.86 \text{ ft/sec}\]
04

Calculate the Vertical Component

Substitute the given values into the formula for the vertical component: \[v_y = 100 \cdot \sin(37^{\circ})\] We need to use the sine of \(37^{\circ}\), which is approximately \(0.6018\). Thus, \[v_y = 100 \cdot 0.6018 = 60.18 \text{ ft/sec}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horizontal Component
To understand the horizontal component of a vector, think about what happens when you move across the ground, parallel to the horizon. This is what we call the horizontal direction. In the context of a projectile like an arrow, the horizontal component measures how far it moves left or right.

When resolving a vector into components, the horizontal component can be calculated using trigonometric functions. Specifically, the horizontal component, denoted here as \(v_x\), is calculated using the cosine function:
  • The formula is \(v_x = v \cdot \cos(\theta)\), where \(v\) is the magnitude of the vector, and \(\theta\) is the angle the vector makes with the horizontal.
  • In the given exercise, \(v = 100\) ft/sec and \(\theta = 37^{\circ}\).
  • Substituting these values in, we get \(v_x = 100 \cdot \cos(37^{\circ})\).
  • Using the approximate value of \(\cos(37^{\circ}) = 0.7986\), the horizontal component \(v_x\) becomes \(79.86\) ft/sec.
Understanding this part helps in figuring out how fast and in what direction the object moves horizontally.
Vertical Component
Now, let's dive into the vertical component, which represents how high or low an object moves. Just like the horizontal component, we calculate the vertical part of a vector using trigonometric methods.

Think of it like watching a ball go up into the sky or fall back to the ground. That's all captured by the vertical component. For this, we often use the sine function:
  • For a vector making an angle \(\theta\) with the horizontal, the vertical component \(v_y\) is found using the formula \(v_y = v \cdot \sin(\theta)\).
  • In our case, the initial speed \(v\) is 100 ft/sec, and the angle \(\theta\) is \(37^{\circ}\).
  • We substitute these into the formula to get \(v_y = 100 \cdot \sin(37^{\circ})\).
  • The approximate value of \(\sin(37^{\circ}) = 0.6018\), so \(v_y\) becomes \(60.18\) ft/sec.
This component reflects the arrow's initial upward velocity and helps us understand how much gravity will affect it as it rises and falls.
Trigonometric Functions
Trigonometric functions, especially sine and cosine, are crucial in decomposing vectors. These functions allow us to translate the direction and magnitude of a vector into easily understandable horizontal and vertical movements.

Here鈥檚 why these functions are essential:
  • Cosine Function: Used to find the horizontal component. It relates the angle \(\theta\) with the ratio of the adjacent side to the hypotenuse in a right triangle. So, \(\cos(\theta)\) helps in estimating how much of the vector is directed horizontally.
  • Sine Function: Used for the vertical component. It describes the angle \(\theta\) in terms of the ratio of the opposite side to the hypotenuse, making it ideal for determining vertical direction.
  • Using these functions helps in breaking down any vector into components that are easier to work with and understand.
  • This is quite useful in physics to predict the behavior of moving objects, like finding out how high or how far it will go given an initial speed and angle.
With the ease of using sine and cosine, determining both the directional components becomes a straightforward and systematic process.

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Most popular questions from this chapter

A UFO is sighted on a direct line between the towns of Batesville and Cave City, sitting stationary in the sky. The towns are \(13 \mathrm{mi}\) apart as the crow flies. A student in Batesville calls a friend in Cave City and both take measurements of the angle of elevation: \(35^{\circ}\) from Batesville and \(42^{\circ}\) from Cave City. Suddenly the UFO zips across the sky at a level altitude heading directly for Cave City, then stops and hovers long enough for an additional measurement from Batesville: \(24^{\circ}\). If the UFO was in motion for \(1.2 \mathrm{sec}\), at what average speed (in mph) did it travel?

Use a half-angle identity to find the value of \(\sin 15^{\circ}\) and \(\cos 75^{\circ}\) in exact form. What do you notice?

Force vectors: For the force vector \(\mathbf{F}\) and vector \(\mathbf{v}\) given, find the amount of work required to move an object along the entire length of \(v_{.}\)Assume force is in pounds and distance in feet. $$\mathbf{F}=\langle 15,10\rangle ; \mathbf{v}=\langle 50,5\rangle$$

Solve each triangle using the law of sines. If the law of sines cannot be used, state why. Draw and label a triangle or label the triangle given before you begin. side \(b=10 \sqrt{3}\) in. $$ \begin{aligned} &\angle A=30^{\circ} \\ &\angle B=60^{\circ} \end{aligned} $$

For each pair of vectors \(u\) and \(v\) given, compute (a) through (d) and illustrate the indicated operations graphically. a. \(\mathbf{u}+\mathbf{v}\) b. \(\mathbf{u}-\mathbf{v}\) c. \(2 \mathbf{u}+1.5 \mathbf{v}\) d. \(\mathbf{u}-2 \mathbf{v}\) $$ \mathbf{u}=\langle-5,-3\rangle ; \mathbf{v}=\langle 6,-4\rangle $$
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