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Chapter 3: Problem 56

A gun shoots a shell into the air with an initial velocity of \(100.0 \mathrm{m} / \mathrm{s}, 60.0^{\circ}\) above the horizontal on level ground. Sketch quantitative graphs of the shell's horizontal and vertical velocity components as functions of time for the complete motion.

Short Answer

Expert verified
Horizontal velocity: constant 50.0 m/s. Vertical velocity: decreases linearly from 86.6 m/s.

Step by step solution

01

Understand the Problem

The shell is fired with an initial velocity of \(100.0\, \mathrm{m/s}\) at an angle of \(60.0^{\circ}\) above the horizontal. We need to sketch graphs of the horizontal and vertical velocity components as functions of time.
02

Determine Horizontal Velocity Component

The horizontal component of velocity, \(v_{x}\), can be found using the formula \(v_{x} = v_0 \cos \theta\), where \(v_0\) is the initial velocity and \(\theta\) is the angle of launch.\[ v_{x} = 100.0 \times \cos(60^{\circ}) = 100.0 \times 0.5 = 50.0\, \mathrm{m/s} \]
03

Graph Horizontal Velocity Component

Since there are no horizontal forces acting on the shell (ignoring air resistance), the horizontal velocity \(v_{x}\) remains constant at \(50.0\, \mathrm{m/s}\) throughout the motion. Sketch a horizontal straight line at \(v_{x} = 50.0\) on a graph with velocity on the y-axis and time on the x-axis.
04

Determine Initial Vertical Velocity Component

The initial vertical component of velocity, \(v_{y0}\), can be found using the formula \(v_{y0} = v_0 \sin \theta\).\[ v_{y0} = 100.0 \times \sin(60^{\circ}) = 100.0 \times \frac{\sqrt{3}}{2} \approx 86.6\, \mathrm{m/s} \]
05

Determine Vertical Velocity as a Function of Time

Vertical velocity changes due to the acceleration due to gravity \(g\), which is approximately \(-9.8\, \mathrm{m/s^2}\). The vertical velocity \(v_{y}(t)\) at any time \(t\) is given by:\[ v_{y}(t) = v_{y0} - gt \]Substitute \(v_{y0} = 86.6\, \mathrm{m/s}\) and \(g = 9.8\, \mathrm{m/s^2}\).\[ v_{y}(t) = 86.6 - 9.8t \]
06

Graph Vertical Velocity Component

Start by plotting \(v_{y0} = 86.6\, \mathrm{m/s}\) at \(t = 0\). The vertical velocity decreases linearly due to gravity, crossing zero at the peak of the trajectory, and then becomes negative as the shell descends. Plot \(v_y(t) = 86.6 - 9.8t\), ensuring it intercepts the x-axis where upward velocity ends.

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Key Concepts

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

Velocity Components in Projectile Motion
In projectile motion, the velocity of an object launched at an angle can be broken down into two main parts: horizontal velocity and vertical velocity. This is crucial because it allows us to analyze the motion in two separate directions.

For our problem, where a shell is fired at a velocity of 100 m/s at an angle of 60掳, we use trigonometric functions:
  • Horizontal Velocity (\(v_{x} = v_{0} \cos \theta \)): This component is calculated using cosine, as it represents the part of the motion parallel to the ground. For a launch angle of 60掳, \( v_{x} = 100 \times 0.5 = 50.0 \, \text{m/s}\).
  • Vertical Velocity (\(v_{y0} = v_{0} \sin \theta \)): This component is calculated using sine, indicating the motion perpendicular to the ground. For 60掳, \( v_{y0} = 100 \times \frac{\sqrt{3}}{2} \approx 86.6 \, \text{m/s} \).
By separating the velocity into these components, we can effectively analyze and predict the path of the projectile.
Acceleration Due to Gravity's Impact
The acceleration due to gravity, denoted as \( g \), is a constant force acting on objects in motion near the Earth's surface. This force acts downward at approximately \(-9.8 \text{ m/s}^2\). It plays a vital role in altering the vertical velocity of a projectile over time.

In our scenario:
  • The horizontal velocity remains unchanged because gravity acts vertically, and no horizontal forces are present (air resistance is ignored).
  • The vertical velocity decreases as the shell rises, reaching zero at the peak (the highest point of the motion), and increases in the downward direction as it falls.
Gravity's influence ensures that the trajectory is parabolic, with the upward path mirroring the downward trajectory.
Understanding Kinematics of a Shell
Kinematics involves the study of motion without considering the forces that cause it. It's crucial for predicting how the shell moves throughout its flight.

For the fired shell, kinematics help us comprehend how its velocity and position change over time. Consider these aspects:
  • Initial conditions: We start with an initial horizontal velocity of 50 m/s and an initial vertical velocity of 86.6 m/s.
  • Velocity equations: The vertical velocity changes with time according to\( v_y(t) = v_{y0} - gt \), where \( v_{y0} = 86.6 \, \text{m/s}\) and \( g = 9.8 \, \text{m/s}^2\).
  • Trajectory: These components combine to form a curved path (parabola) as the shell moves through the air.
This understanding allows us to graphically represent how the shell's velocity changes over time and predict where it will land.
Graphical Representation of Shell's Motion
Graphical representation is a powerful tool in visualizing the motion of the projectile and understanding the changes in velocity over time. Let's create two graphs鈥攐ne for horizontal and one for vertical velocity.

Here's how:
  • Horizontal Velocity: Plot a horizontal line at 50 m/s on the graph, indicating that the horizontal velocity remains constant as there are no forces affecting it. Time is on the x-axis, and velocity is on the y-axis.
  • Vertical Velocity: Start the graph at 86.6 m/s (initial vertical velocity) at \( t = 0 \). The slope of the line is negative, with a decrease of 9.8 m/s each second due to gravity. This line dips to zero at the peak of the trajectory before continuing negatively as the shell falls.
Single lines on these graphs reveal much about the motion, helping students visualize and comprehend the projectile's path.

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Most popular questions from this chapter

\(\bullet\) Leaping the river, A \(10,000 \mathrm{N}\) car comes to a bridge during a storm and finds the bridge washed out. The 650 \(\mathrm{N}\) driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 21.3 \(\mathrm{m}\) above the river, while the opposite side is a mere 1.80 \(\mathrm{m}\) above the river. The river itself is a raging torrent 61.0 \(\mathrm{m}\) wide. (a) How fast should the car be traveling just as it leaves the cliff in order to clear the river and land safely on the opposite side? (b) What is the speed of the car just before it lands safely on the other side?

\bullet A cart carrying a vertical missile launcher moves horizontally at a constant velocity of 30.0 \(\mathrm{m} / \mathrm{s}\) to the right. It launches a rocket vertically upward. The missile has an initial vertical velocity of 40.0 \(\mathrm{m} / \mathrm{s}\) relative to the cart. (a) How high does the rocket go? (b) How far does the cart travel while the rocket is in the air? (c) Where does the rocket land relative to the cart?

\(\bullet\) Bird migration. Canadian geese migrate essentially along a north- south direction for well over a thousand kilometers in some cases, traveling at speeds up to about 100 \(\mathrm{km} / \mathrm{h}\) . If one such bird is flying at 100 \(\mathrm{km} / \mathrm{h}\) relative to the air, but there is a 40 \(\mathrm{km} / \mathrm{h}\) wind blowing from west to east, (a) at what angle relative to the north-south direction should this bird head so that it will be traveling directly southward relative to the ground? (b) How long will it take the bird to cover a ground distance of 500 \(\mathrm{km}\) from north to south? (Note: Even on cloudy nights, many birds can navigate using the earth's magnetic field to fix the north-south direction.)

\bulletA test rocket is launched by accelerating it along a \(200.0-\mathrm{m}\) incline at 1.25 \(\mathrm{m} / \mathrm{s}^{2}\) starting from rest at point \(A\) (Figure \(3.40 . )\) The incline rises at \(35.0^{\circ}\) above the horizontal,and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored). Find (a) the maximum height above the ground that the rocket reaches, and (b) the greatest horizontal range of the rocket beyond point \(A .\)

At an air show, a jet plane has velocity components \(v_{x}=\) 625 \(\mathrm{km} / \mathrm{h}\) and \(v_{y}=415 \mathrm{km} / \mathrm{h}\) at time 3.85 \(\mathrm{s}\) and \(v_{x}=838 \mathrm{km} / \mathrm{h}\) and \(v_{y}=365 \mathrm{km} / \mathrm{h}\) at time 6.52 s. For this time interval, find(a) the \(x\) and \(y\) components of the plane's average acceleration and (b) the magnitude and direction of its average acceleration.
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