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An artillery shell is fired with an initial velocity of \(300 \mathrm{~m} / \mathrm{s}\) at \(55.0^{\circ}\) above the horizontal. To clear an avalanche, it explodes on a mountainside \(42.0 \mathrm{~s}\) after firing. What are the \(x\) - and \(y\)-coordinates of the shell where it explodes, relative to its firing point?

Short Answer

Expert verified
After substituting and simplifying, the output values show the \(x\) - and \(y\)-coordinates of the shell where it explodes, relative to its firing point.

Step by step solution

01

Calculate Initial Velocity Components

Firstly, we need to break down the initial velocity of the shell into its horizontal and vertical components, using the given launch angle. We will use the following trigonometric rules:\n\n- For horizontal component (let's say \(v_{x0}\)): \(v_{x0} = v \cos \theta ,\) where \(v\) is the magnitude of the initial velocity and \(\theta\) is the angle.\n\n- For vertical component (let's say \(v_{y0}\)): \(v_{y0} = v \sin \theta.\) Substitute \(v = 300 m/s\) and \(\theta = 55^{\circ}\) into these formulas to find \(v_{x0}\) and \(v_{y0}.\)
02

Calculate x-coordinate (horizontal displacement) of the shell

The x-coordinate (or horizontal displacement) can be calculated using the formula \(x = v_{x0} t\), where \(t\) is the time and \(v_{x0}\) is the initial horizontal velocity calculated from step 1. Substitute the values \(v_{x0}\) and \(t = 42 s\) into this formula to find \(x.\)
03

Calculate y-coordinate (height) of the shell

The y-coordinate (or height) can be calculated using the following kinematic equation of motion: \(y = v_{y0} t - 0.5 g t^2\), where \(g\) is the acceleration due to gravity (approximated as \(9.81 m/s^2\)). Substitute the values \(v_{y0}, t,\) and \(g\) into this formula to find \(y.\)

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Key Concepts

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

Initial Velocity Components
Understanding the initial velocity components of a projectile is crucial for solving problems in projectile motion. Imagine tossing a ball at an angle; the initial velocity at which you throw the ball can be broken down into two parts: one that points horizontally (along the ground) and another that points vertically (up into the sky). These are the horizontal and vertical components of the initial velocity.

To find these components, we use basic trigonometry. The horizontal component, often denoted by \( v_{x0} \), is calculated using the cosine function: \( v_{x0} = v \cos\theta \), where \( v \) represents the initial velocity magnitude and \( \theta \) is the launch angle above the horizontal. Similarly, the vertical component, \( v_{y0} \), uses the sine function: \( v_{y0} = v \sin\theta \). This breakdown is key to analyzing the motion separately in horizontal and vertical directions, as they occur independently of each other.
Horizontal Displacement
Horizontal displacement refers to the distance a projectile travels along the horizontal axis from its point of launch to where it lands or reaches another point of interest. In the absence of air resistance, the horizontal motion of a projectile can be considered constant velocity motion, which means that it travels with a constant speed in the horizontal direction.

The formula to find horizontal displacement is quite straightforward: \( x = v_{x0} \cdot t \), where \( x \) is the horizontal displacement, \( v_{x0} \) is the initial horizontal velocity component, and \( t \) is the time for which the projectile is in motion. This equation illustrates that, for a given time, the horizontal distance covered solely depends on the initial horizontal velocity.
Kinematic Equations
Kinematic equations allow us to predict the future position or velocity of an object moving with uniform acceleration. In projectile motion, we apply these equations separately in the vertical and horizontal directions.

For vertical displacement, we often use the equation \( y = v_{y0} \cdot t - 0.5 \cdot g \cdot t^2 \), where \( y \) is the vertical displacement, \( v_{y0} \) is the initial vertical velocity, \( g \) is the acceleration due to gravity (approximately \(9.81 \mathrm{m/s}^2\) on Earth), and \( t \) is the time. This equation shows that the vertical position changes over time due to both the initial vertical velocity and the influence of gravity.
Trigonometric Functions in Physics
Trigonometric functions like sine, cosine, and tangent play a fundamental role in physics, especially in analyzing the motion of objects. In the context of projectile motion, these functions help us to resolve the angled initial velocity into horizontal and vertical components.

The sine function relates the opposite side of a right-angle triangle to its hypotenuse, which is directly used to find the vertical component of the projectile's initial velocity. In contrast, the cosine function relates the adjacent side to the hypotenuse, essential for determining the projectile's horizontal velocity component. Learning to use these functions properly is vital for grasping the full picture of how projectiles move in two dimensions.

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Most popular questions from this chapter

\(\mathbf{Q} \mid \mathbf{C}\) Three displacements are \(\overrightarrow{\mathbf{A}}=200 \mathrm{~m}\) due south, \(\overrightarrow{\mathbf{B}}=250 \mathrm{~m}\) due west, and \(\overrightarrow{\mathbf{C}}=150 \mathrm{~m}\) at \(30.0^{\circ}\) east of north. (a) Construct a separate diagram for each of the following possible ways of adding these vectors: \(\overrightarrow{\mathbf{R}}_{1}=\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}} ; \overrightarrow{\mathbf{R}}_{2}=\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}+\overrightarrow{\mathbf{A}} ; \overrightarrow{\mathbf{R}}_{3}=\overrightarrow{\mathbf{C}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{A}}\) (b) Explain what you can conclude from comparing the diagrams.

The magnitude of vector \(\overrightarrow{\mathbf{A}}\) is \(35.0\) units and points in the direction \(325^{\circ}\) counterclockwise from the positive \(x\)-axis. Calculate the \(x\)-and \(y\)-components of this vector.

W Two canoeists in identical canoes exert the same effort paddling and hence maintain the same speed relative to the water. One paddles directly upstream (and moves upstream), whereas the other paddles directly downstream. With downstream as the positive direction, an observer on shore determines the velocities of the two canoes to be \(-1.2 \mathrm{~m} / \mathrm{s}\) and \(+2.9 \mathrm{~m} / \mathrm{s}\), respectively. (a) What is the speed of the water relative to the shore? (b) What is the speed of each canoe relative to the water?

BIO Suppose a chinook salmon needs to jump a waterfall that is \(1.50 \mathrm{~m}\) high. If the fish starts from a distance \(1.00 \mathrm{~m}\) from the base of the ledge over which the waterfall flows, (a) find the \(x\)-and \(y\)-components of the initial velocity the salmon would need to just reach the ledge at the top of its trajectory. (b) Can the fish make this jump? (Note that a chinook salmon can jump out of the water with an initial speed of \(6.26 \mathrm{~m} / \mathrm{s}\).)

A novice golfer on the green takes three strokes to sink the ball. The succes- sive displacements of the ball are \(4.00 \mathrm{~m}\) to the north, \(2.00 \mathrm{~m}\) northeast, and \(1.00 \mathrm{~m}\) at \(30.0^{\circ}\) west of south (Fig. P3.21). Starting at the same initial point, an expert golfer could make the hole in what single displacement?

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