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Magazine Chapter 4: Problem 31
A cork shoots out of a champagne bottle at an angle of \(35.0^{\circ}\) above the horizontal. If the cork travels a horizontal distance of \(1.30 \mathrm{m}\) in \(1.25 \mathrm{s},\) what was its initial speed?
Short Answer
Expert verified
The initial speed of the cork was approximately \(1.27 \text{ m/s}\).
Step by step solution
01
Understand the Problem
We need to find the initial speed of the cork that moves in projectile motion. We know the angle of projection is \(35.0^{\circ}\), the horizontal distance traveled is \(1.30 \text{ m}\), and the time of flight is \(1.25 \text{ s}\).
02
Horizontal Motion Formula
For horizontal motion at constant velocity, use the formula \( R = v_{0x} \times t \). Here \( R = 1.30 \text{ m} \) and \( t = 1.25 \text{ s} \), where \( v_{0x} = v_0 \cos \theta \).
03
Solve for Horizontal Component of Velocity
Rearrange the horizontal motion formula to solve for horizontal component \( v_{0x} \), \( v_{0x} = \frac{R}{t} \). Substitute \( R = 1.30 \text{ m} \) and \( t = 1.25 \text{ s} \): \( v_{0x} = \frac{1.30}{1.25} = 1.04 \text{ m/s} \)
04
Relate Initial Speed with Components
The initial velocity \( v_0 \) has components \( v_{0x} = v_0 \cos \theta \) and \( v_{0y} = v_0 \sin \theta \). We have \( v_{0x} = 1.04 \text{ m/s} \) and \( \cos \theta = \cos 35.0^{\circ} \).
05
Calculate Initial Speed
Utilize the relation \( v_0 \cos \theta = v_{0x} \). Rearrange to find \( v_0 = \frac{v_{0x}}{\cos \theta} \). Substitute \( v_{0x} = 1.04 \text{ m/s} \) and \( \cos 35.0^{\circ} \approx 0.819 \): \( v_0 = \frac{1.04}{0.819} \approx 1.27 \text{ m/s} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Velocity Calculation
When analyzing projectile motion, calculating the initial velocity is crucial as it determines how the object moves. In this scenario, we need to know the speed at which the cork was launched from the champagne bottle.
We are given the horizontal distance (or range) the cork travels, which is **1.30 meters**, and the time of flight, which is **1.25 seconds**. The angle of projection is provided as **35 degrees** above the horizontal. The initial velocity \(v_0\) is split into two components based on this angle: horizontal \(v_{0x}\) and vertical \(v_{0y}\).
To find \(v_{0x}\), we use the formula for constant horizontal motion, \(v_{0x} = \frac{R}{t}\), where \(R\) is the range, and \(t\) is the time. Substituting the known values gives us a horizontal velocity component of **1.04 m/s**. Using the relationship \(v_{0x} = v_0 \cos \theta\), we rearrange to solve for the initial speed and get approximately \(1.27 m/s\).
We are given the horizontal distance (or range) the cork travels, which is **1.30 meters**, and the time of flight, which is **1.25 seconds**. The angle of projection is provided as **35 degrees** above the horizontal. The initial velocity \(v_0\) is split into two components based on this angle: horizontal \(v_{0x}\) and vertical \(v_{0y}\).
To find \(v_{0x}\), we use the formula for constant horizontal motion, \(v_{0x} = \frac{R}{t}\), where \(R\) is the range, and \(t\) is the time. Substituting the known values gives us a horizontal velocity component of **1.04 m/s**. Using the relationship \(v_{0x} = v_0 \cos \theta\), we rearrange to solve for the initial speed and get approximately \(1.27 m/s\).
Horizontal Motion
In projectile motion, understanding horizontal motion separately from vertical motion can simplify calculations. Horizontal motion refers to how an object travels along a straight path parallel to the ground.
In this particular problem, the horizontal motion is analyzed by using the formula \(R = v_{0x} \times t\), where \(R\) is the horizontal distance traveled, \(v_{0x}\) is the horizontal component of initial velocity, and \(t\) is the time of flight. For the cork, we found \(v_{0x}\) by isolating it in the equation, giving us **1.04 m/s**.
This velocity component remains constant because gravity does not affect horizontal motion, assuming there's no air resistance. Therefore, understanding this aspect helps bridge how initial velocity breakdown facilitates overall motion within projectile dynamics.
In this particular problem, the horizontal motion is analyzed by using the formula \(R = v_{0x} \times t\), where \(R\) is the horizontal distance traveled, \(v_{0x}\) is the horizontal component of initial velocity, and \(t\) is the time of flight. For the cork, we found \(v_{0x}\) by isolating it in the equation, giving us **1.04 m/s**.
This velocity component remains constant because gravity does not affect horizontal motion, assuming there's no air resistance. Therefore, understanding this aspect helps bridge how initial velocity breakdown facilitates overall motion within projectile dynamics.
Angle of Projection
The angle of projection significantly affects how a projectile travels through space. It's the angle between the initial velocity vector and the horizontal axis.
For this exercise, the cork's angle of projection is **35 degrees** above the horizontal. This angle will determine how the initial velocity is split into horizontal and vertical components. Specifically, \(v_0 \cos \theta\) gives the horizontal component, while \(v_0 \sin \theta\) gives the vertical component.
Understanding the angle of projection is crucial because it impacts both the range and maximum height of the projectile. Generally, for any projectile, the angle will dictate the balance between height and distance achieved, which are inversely proportional based on the angle.
For this exercise, the cork's angle of projection is **35 degrees** above the horizontal. This angle will determine how the initial velocity is split into horizontal and vertical components. Specifically, \(v_0 \cos \theta\) gives the horizontal component, while \(v_0 \sin \theta\) gives the vertical component.
Understanding the angle of projection is crucial because it impacts both the range and maximum height of the projectile. Generally, for any projectile, the angle will dictate the balance between height and distance achieved, which are inversely proportional based on the angle.
Time of Flight
Time of flight in projectile motion refers to the total time an object spends in the air. It's a critical variable when discussing how far a projectile travels.
In this problem, the cork remains airborne for **1.25 seconds**. This time can be used with the horizontal motion equation to determine components of initial velocity, as seen in previous sections.
During projectile motion, the time of flight is influenced by the initial velocity and launch angle. In ideal scenarios without air resistance, the time to reach the peak is equal to the time it takes to descend back to the same horizontal level, but this exercise primarily focuses on its use to calculate velocity components.
In this problem, the cork remains airborne for **1.25 seconds**. This time can be used with the horizontal motion equation to determine components of initial velocity, as seen in previous sections.
During projectile motion, the time of flight is influenced by the initial velocity and launch angle. In ideal scenarios without air resistance, the time to reach the peak is equal to the time it takes to descend back to the same horizontal level, but this exercise primarily focuses on its use to calculate velocity components.
