Vector Problem: Solving Displacement & Speed of Contact

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SUMMARY

The discussion focuses on solving a vector problem related to displacement and speed of contact with the ground. Participants emphasize the importance of using the equation of motion and vector components, specifically highlighting the horizontal nature of the velocity vector, denoted as ##\vec{V}=U\hat{V}##. The conversation suggests that the final displacement point should be at coordinates (a, b, 0), indicating a two-dimensional approach to the problem. The mention of quantum mechanics (QM) is noted, but it is clarified that the problem does not require QM concepts for resolution.

PREREQUISITES
  • Understanding of vector mathematics and operations, including dot and cross products.
  • Familiarity with the equations of motion in physics.
  • Knowledge of coordinate systems, particularly two-dimensional Cartesian coordinates.
  • Basic concepts of quantum mechanics (QM) for context, though not essential for solving the problem.
NEXT STEPS
  • Study vector operations, focusing on dot and cross products in physics applications.
  • Review the equations of motion and their applications in two-dimensional problems.
  • Explore the implications of horizontal and vertical components in vector analysis.
  • Investigate the relationship between displacement and speed in kinematic equations.
USEFUL FOR

Students and professionals in physics, particularly those dealing with vector analysis, kinematics, and motion problems. This discussion is beneficial for anyone looking to enhance their understanding of displacement and speed in two-dimensional contexts.

dreamer_asot
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Hello, my problem is attached as a picture. Could you give me some guidelines on how to approach the problem? (i know the given formulas are derived using QM (probably), and I'm not "scared" from them, i just need to know where to do a cross or a dot product, and maybe how to approach the last part of the problem - the speed of contact with the ground. This would probably mean that the vector displacement finish point will be at (a, b, 0).)
 

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I don't see any QM in there.
You have the equation of motion along with clues to ##\vec{V}## and ##\vec{A}## - why not use this?

It may help if you pick a direction for U - if it has to be general, then ##\vec{V}=U\hat{V}## which will have zero y component (because it is "horizontal").

Notice the the y-axis is "up".
 

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