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Chapter 3: Problem 16

On level ground a shell is fired with an initial velocity of 40.0 m/s at 60.0\(^\circ\) above the horizontal and feels no appreciable air resistance. (a) Find the horizontal and vertical components of the shell's initial velocity. (b) How long does it take the shell to reach its highest point? (c) Find its maximum height above the ground. (d) How far from its firing point does the shell land? (e) At its highest point, find the horizontal and vertical components of its acceleration and velocity.

Short Answer

Expert verified
The shell lands 141.2 m away. The maximum height is 61.06 m.

Step by step solution

01

Break down the initial velocity into components

To find the horizontal and vertical components of the initial velocity, use trigonometric functions. Given that the initial velocity \( v_0 \) is 40 m/s and the angle \( \theta = 60^\circ \): - The horizontal component \( v_{0x} = v_0 \cos(\theta) = 40 \cos(60^\circ) = 40 \times 0.5 = 20 \text{ m/s} \).- The vertical component \( v_{0y} = v_0 \sin(\theta) = 40 \sin(60^\circ) = 40 \times 0.866 = 34.64 \text{ m/s} \).
02

Calculate the time to reach the highest point

The time to reach the highest point occurs when the vertical velocity component becomes zero. Using the equation \( v_{y} = v_{0y} - gt \), where \( g \) is the acceleration due to gravity (9.81 m/s虏):\[ 0 = 34.64 - 9.81t \]Solving for \( t \):\[ t = \frac{34.64}{9.81} \approx 3.53 \text{ seconds} \]
03

Determine the maximum height

Using the formula for vertical motion \( h = v_{0y}t - \frac{1}{2}gt^2 \) with \( t = 3.53 \text{ s} \):\[ h = 34.64 \times 3.53 - \frac{1}{2} \times 9.81 \times (3.53)^2 \]\[ h = 122.25 - 61.19 \approx 61.06 \text{ meters} \]
04

Find the range of the shell

The range \( R \) of the projectile is found using \( R = v_{0x} \times 2t_{up} \) since total time in the air is twice the time to the highest point:\[ R = 20 \times 2 \times 3.53 = 20 \times 7.06 = 141.2 \text{ meters} \]
05

Analyze horizontal and vertical components at the highest point

At its highest point, the shell's vertical velocity component \( v_y \) becomes 0 m/s, while the horizontal velocity component remains unchanged at 20 m/s, given no air resistance. Both components of acceleration remain \( a_x = 0 \text{ m/s}^2 \) (no horizontal forces) and \( a_y = -9.81 \text{ m/s}^2 \) (due to gravity).

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Key Concepts

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

Kinematics
Kinematics is the branch of physics that describes the motion of objects without considering the forces that cause the motion. It deals with concepts such as displacement, velocity, and acceleration. In the context of projectile motion, kinematics helps us analyze the path of an object under the influence of gravity solely.

When studying the motion of a projectile like a shell fired from a cannon, we are interested in determining various parameters:
  • Initial Velocity: The velocity at which the projectile is launched.
  • Components of Velocity: Breakdown of this velocity into horizontal and vertical elements.
  • Time of Flight: The total time for which the projectile remains in motion.
  • Maximum Height: The highest vertical point reached by the projectile.
  • Range: The horizontal distance covered by the projectile.
Understanding these components gives a comprehensive view of the projectile's trajectory and behavior.
Trigonometry in Physics
Trigonometry plays a crucial role in physics, especially in breaking down forces and motions into their respective components. In projectile motion, we often need to resolve the initial velocity into horizontal and vertical components to analyze the motion accurately.

By using basic trigonometric functions:
  • Cosine ( \(\cos\) ): Helps determine the horizontal component of velocity. For a given initial velocity \(v_0\) at an angle \(\theta\), the horizontal component is calculated as \(v_{0x} = v_0 \cos(\theta)\).
  • Sine ( \(\sin\) ): Useful for calculating the vertical component of velocity. It is given by \(v_{0y} = v_0 \sin(\theta)\).
Applying these functions allows us to transform a single vector into manageable segments, making it easier to solve complex motion problems.
Components of Velocity
In projectile motion, understanding the components of velocity is essential for determining how an object travels through space. The initial velocity of the shell can be dissected into two main components:

  • Horizontal Component ( \(v_{0x} \)): Governs the constant horizontal motion. Since there is no air resistance in this scenario, this velocity remains unchanged throughout the flight. Calculated as \(v_{0x} = 40 \times \cos(60^\circ) = 20 \text{ m/s}\).
  • Vertical Component ( \(v_{0y} \)): Affected by gravity, this component changes over time. The initial vertical component is calculated as \(v_{0y} = 40 \times \sin(60^\circ) = 34.64 \text{ m/s}\).
The interplay between these components determines the projectile's path and maximum height.
Acceleration due to Gravity
Acceleration due to gravity is a vital factor in projectile motion. It is a constant force pulling objects towards the Earth's surface at approximately \(9.81 \text{ m/s}^2\). This acceleration affects only the vertical component of a projectile's velocity and is independent of the object's horizontal motion.

Because of gravity:
  • The vertical component of velocity decreases as the object rises, reaching zero at the apex of the projectile's path.
  • After reaching maximum height, the vertical velocity increases in the downward direction as the object falls back to the ground.
Understanding this constant acceleration is crucial for predicting how long the projectile will be in the air and how high it will go.

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Most popular questions from this chapter

A jet plane is flying at a constant altitude. At time \(t_1\) = 0, it has components of velocity \(v_x\) = 90 m/s, \(v_y\) = 110 m/s. At time \(t_2\) = 30.0 s, the components are \(v_x\) = -170 m/s, \(v_y\) = 40 m/s. (a) Sketch the velocity vectors at \(t_1\) and \(t_2\). How do these two vectors differ? For this time interval calculate (b) the components of the average acceleration, and (c) the magnitude and direction of the average acceleration.

An airplane pilot sets a compass course due west and maintains an airspeed of 220 km/h. After flying for 0.500 h, she finds herself over a town 120 km west and 20 km south of her starting point. (a) Find the wind velocity (magnitude and direction). (b) If the wind velocity is 40 km/h due south, in what direction should the pilot set her course to travel due west? Use the same airspeed of 220 km/h.

A dog running in an open field has components of velocity \(v_x\) = 2.6 m/s and \(v_y\) = -1.8 m/s at \(t_1\) = 10.0 s. For the time interval from \(t_1\) = 10.0 s to \(t_2\) = 20.0 s, the average acceleration of the dog has magnitude 0.45 m/s\(^2\) and direction 31.0\(^\circ\) measured from the +\(x\)-axis toward the +\(y\)-axis. At \(t_2\) = 20.0 s, (a) what are the \(x\)- and \(y\)-components of the dog's velocity? (b) What are the magnitude and direction of the dog's velocity? (c) Sketch the velocity vectors at \(t_1\) and \(t_2\). How do these two vectors differ?

A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.480 s. Ignore air resistance. Find (a) the height of the tabletop above the floor; (b) the horizontal distance from the edge of the table to the point where the book strikes the floor; (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before the book reaches the floor. (d) Draw \(x-t, y-t, v_x-t\), and \(v_y-t\) graphs for the motion.

A baseball thrown at an angle of 60.0\(^{\circ}\) above the horizontal strikes a building 18.0 m away at a point 8.00 m above the point from which it is thrown. Ignore air resistance. (a) Find the magnitude of the ball's initial velocity (the velocity with which the ball is thrown). (b) Find the magnitude and direction of the velocity of the ball just before it strikes the building.
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