Win the Prize Projectile Motion Problem

that physics tutor
16 Jun 202310:32
EducationalLearning
32 Likes 10 Comments

TLDRIn this educational video, the host tackles problem 3.19 from Young and Friedman's 'University Physics' textbook, which involves projectile motion. The scenario is a carnival game where a quarter must be tossed into a dish 2.1 meters away. The host calculates the shelf's height and the coin's vertical velocity before landing, given an initial velocity of 6.4 m/s at a 60-degree angle. The solution uses kinematic equations, ignoring air resistance, and finds the shelf's height to be 1.527 meters and the vertical velocity component to be -0.89 m/s, just before the coin lands. The video is designed to help viewers understand the physics of projectile motion and is suitable for students and learners.

Takeaways
  • 🎯 The problem is from Young and Friedman's 'University Physics' textbook, specifically problem 3.19 titled 'Win the Prize'.
  • 🔍 The scenario involves tossing a quarter into a dish at a carnival booth to win a prize, with the dish located 2.1 meters horizontally from the point of release.
  • 🚀 The coin is tossed with an initial velocity of 6.4 m/s at a 60-degree angle above the horizontal, and air resistance is ignored.
  • 📏 The task is to find the height of the shelf above the point where the quarter is released and the vertical component of the coin's velocity just before it lands.
  • 📚 The problem provides a diagram, which helps visualize the situation and the given parameters.
  • 📉 The horizontal component of the initial velocity (vix) is calculated using the cosine of the angle, resulting in 6.4 * cos(60).
  • 📈 The time of flight is found by dividing the horizontal distance by the horizontal component of the initial velocity, yielding 2.1 / (6.4 * cos(60)) seconds.
  • ⏱ The calculated time of flight is approximately 0.65625 seconds, which is used to find the vertical displacement.
  • 📊 The vertical displacement (dy) is calculated using the kinematic equation dy = viy * t + 0.5 * a * t^2, where a is the acceleration due to gravity (-9.8 m/s^2).
  • 📐 The height of the shelf (dy) is found to be approximately 1.527 meters above the point where the quarter is released.
  • 🌟 For part B, the vertical component of the coin's velocity just before landing (vfy) is calculated using the equation vfy = viy + a * t, resulting in -0.89 m/s.
Q & A

  • What is the main objective of the problem presented in the video?

    -The main objective of the problem is to determine the height of the shelf above the point where the quarter leaves the hand and to find the vertical component of the velocity of the quarter just before it lands in the dish.

  • What textbook does the problem come from?

    -The problem comes from Young and Friedman's 'University Physics' textbook.

  • What is the initial velocity of the quarter when tossed?

    -The initial velocity of the quarter when tossed is 6.4 meters per second.

  • At what angle is the quarter tossed with respect to the horizontal?

    -The quarter is tossed at a 60-degree angle above the horizontal.

  • What is the horizontal distance from the point where the quarter is released to the dish?

    -The horizontal distance from the point where the quarter is released to the dish is 2.1 meters.

  • What is the first step in solving the problem?

    -The first step in solving the problem is to calculate the time of flight using the horizontal motion equations.

  • How is the time of flight calculated in the video?

    -The time of flight is calculated by dividing the horizontal distance (2.1 meters) by the horizontal component of the initial velocity (6.4 m/s * cos(60 degrees)).

  • What is the formula used to find the vertical height (dy) of the shelf?

    -The formula used to find the vertical height (dy) is dy = viy * t + 0.5 * a * t^2, where viy is the initial vertical velocity, a is the acceleration due to gravity, and t is the time of flight.

  • What is the calculated height of the shelf above the point where the quarter leaves the hand?

    -The calculated height of the shelf is 1.527 meters above the point where the quarter leaves the hand.

  • What is the vertical component of the velocity of the quarter just before it lands in the dish?

    -The vertical component of the velocity of the quarter just before it lands in the dish is -0.89 meters per second.

  • Why is the formula vfy = viy + a*t preferred over vfy^2 = viy^2 + 2*a*dy in this context?

    -The formula vfy = viy + a*t is preferred because it does not involve squaring the velocities, which can lead to a loss of information regarding the direction of motion.

  • What does the speaker suggest doing if the viewer finds the video helpful?

    -The speaker suggests liking, subscribing, and sharing the video with friends or classmates if the viewer finds it helpful.

Outlines
00:00
🎯 Projectile Motion Problem Introduction

This paragraph introduces a physics problem from Young and Friedman's 'University Physics' textbook, specifically problem 3.19 titled 'Win the Prize in a Carnival Booth.' The problem involves tossing a quarter into a small dish to win a prize. The dish is located 2.1 meters horizontally from the point where the quarter is released, and the quarter is tossed with an initial velocity of 6.4 m/s at a 60-degree angle above the horizontal. The goal is to determine the height of the shelf and the vertical component of the quarter's velocity just before it lands in the dish, assuming no air resistance. The paragraph outlines the given information and sets up the problem by discussing the initial velocity components and the horizontal distance to the dish.

05:01
📚 Solving for the Height of the Shelf

The speaker begins by calculating the time of flight for the quarter using the horizontal component of its initial velocity and the horizontal distance to the dish. They find the time to be approximately 0.65625 seconds. With the time determined, they proceed to use kinematic equations to find the vertical height (d_y) the quarter travels before landing in the dish. The vertical component of the initial velocity is calculated, and the kinematic equation d_y = vi_y * t + 0.5 * a * t^2 is applied, where a is the acceleration due to gravity (-9.8 m/s^2). The calculated height of the shelf is 1.527 meters above the point where the quarter is released.

10:03
🚀 Calculating the Vertical Velocity Component Before Landing

In this paragraph, the focus shifts to solving for the vertical component of the quarter's velocity (vf_y) just before it lands in the dish. The speaker uses the kinematic equation vf_y = vi_y + a * t, avoiding the squared version of the equation to preserve direction information. By plugging in the values for the initial vertical velocity component, acceleration due to gravity, and the time of flight, they calculate the final vertical velocity to be approximately -0.89 m/s, indicating a downward velocity as the quarter lands. The paragraph concludes with a note on the difficulty level of the problem, described as a two-dot problem, and a reminder to the viewers about the potential for different numbering in various editions of the textbook.

📢 Conclusion and Viewer Engagement

The final paragraph wraps up the problem-solving session by summarizing the solutions found for both parts of the problem: the height of the shelf (1.527 meters) and the vertical component of the quarter's velocity just before landing (-0.89 m/s). The speaker encourages viewers to ask questions or provide feedback via comments or email if they find the explanation helpful. They also remind viewers to like, subscribe, and share the video with friends or classmates, emphasizing the value of the content for educational purposes. The video concludes with a thank you and a goodbye, setting the stage for future sessions.

Mindmap
Keywords
💡Projectile Motion
Projectile motion is a form of motion experienced by an object or particle that is thrown near the Earth's surface and moves along a curved path under the action of gravity only. In the video, the concept of projectile motion is central to solving the problem of how to toss a quarter into a dish at a carnival booth. The coin's motion is analyzed in both horizontal and vertical components, which is a key aspect of understanding projectile motion.
💡Initial Velocity
Initial velocity, often denoted as 'v_i', is the velocity of an object at the start of its motion. In the context of the video, the initial velocity of the quarter is given as 6.4 meters per second at an angle of 60 degrees above the horizontal. This value is critical in determining the coin's trajectory and is used to calculate both the horizontal and vertical components of its motion.
💡Horizontal Distance
Horizontal distance refers to the distance an object travels in the horizontal plane. In the video script, the horizontal distance from the point where the quarter is released to the dish is specified as 2.1 meters. This distance is used in conjunction with the horizontal component of the initial velocity to calculate the time of flight.
💡Angle of Projection
The angle of projection is the angle at which a projectile is launched into the air. In the script, the quarter is tossed at a 60-degree angle above the horizontal. This angle is essential for determining the trajectory of the quarter and is used to calculate the vertical and horizontal components of the initial velocity.
💡Time of Flight
Time of flight is the time taken by an object in projectile motion to reach its landing point. The video script involves calculating the time of flight for the quarter to travel the 2.1-meter horizontal distance. This is done by dividing the horizontal distance by the horizontal component of the initial velocity.
💡Vertical Component
The vertical component of a vector quantity, such as velocity or displacement, refers to the part of the quantity that acts in the upward or downward direction. In the video, the vertical component of the initial velocity (v_iy) and the final velocity (vf_y) are calculated to determine the height of the shelf and the coin's velocity just before landing, respectively.
💡Acceleration Due to Gravity
Acceleration due to gravity is the acceleration that an object experiences when it falls freely under the influence of Earth's gravity. It is denoted as 'g' and is approximately -9.8 m/s^2 downwards. In the video, this acceleration is used in the kinematic equations to calculate the vertical displacement and final vertical velocity of the quarter.
💡Kinematic Equations
Kinematic equations are mathematical formulas that relate an object's displacement, initial velocity, final velocity, acceleration, and time. In the video, kinematic equations are used to solve for unknowns such as the height of the shelf (d_y) and the final vertical velocity (vf_y) of the quarter.
💡Carnival Booth
A carnival booth is a stall or kiosk at a fair or carnival where games or activities are conducted for entertainment and prizes. The video script presents a problem related to winning a prize at a carnival booth by tossing a quarter into a dish, which serves as a practical application of the physics concepts discussed.
💡Problem Solving
Problem solving in the context of the video refers to the process of applying physics principles to find solutions to practical problems. The video demonstrates problem-solving techniques by guiding the viewer through the steps needed to determine the height of the shelf and the final vertical velocity of the quarter in a carnival booth game.
💡Diagram
A diagram is a visual representation used to illustrate the components of a problem or the relationships between different elements. Although the video script mentions that a diagram is provided in the textbook, it also explains the problem conceptually without relying on the diagram, helping viewers understand the physics concepts involved.
Highlights

Introduction to problem 3.19 from Young and Friedman's University Physics textbook.

Problem involves winning a prize at a carnival booth by tossing a quarter into a dish.

The dish is located 2.1 meters horizontally from the point where the quarter is released.

The coin is tossed with an initial velocity of 6.4 m/s at a 60-degree angle above the horizontal.

Air resistance is ignored in the problem.

Objective is to find the height of the shelf above the release point.

The problem is broken down into two parts: finding the height (Part A) and the vertical velocity component before landing (Part B).

Uses the given diagram to understand the problem setup.

Calculates the horizontal component of the initial velocity using cosine of 60 degrees.

Determines the time of flight by dividing the horizontal distance by the horizontal velocity.

Finds the time of flight to be 0.65625 seconds.

Uses kinematic equations to find the vertical height (d_y) of the shelf.

Calculates the vertical height to be 1.527 meters above the release point.

Problem difficulty is rated as two blue dots, indicating a moderate level of challenge.

For Part B, the vertical component of the velocity (vf_y) is calculated using the time found earlier.

The vertical velocity component just before landing is found to be -0.89 m/s.

Encourages viewers to ask questions, like, subscribe, and share the video if it was helpful.

Transcripts

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Physics ProblemProjectile MotionEducationalUniversity PhysicsProblem SolvingCarnival BoothKinematic EquationsVelocity AnalysisYoung and FriedmanEducational Content