Is the Angle of a Projectile at 1/4 of its Flight the Same as When it is Fired?

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Homework Help Overview

The discussion revolves around the analysis of projectile motion, specifically examining the angle of a projectile at one-fourth of its flight time compared to its launch angle. The problem requires an algebraic approach without numerical values, focusing on the relationship between the projectile's velocity components and the angles involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the angle of the projectile at different points in its flight and question whether the angle at one-fourth of the flight is the same as the launch angle. There are discussions about the definitions of slope and tangent in relation to the projectile's trajectory, as well as the components of velocity at specific times.

Discussion Status

The discussion is ongoing, with participants providing insights into the nature of projectile motion and the mathematical relationships involved. Some participants have suggested clarifying the distinction between initial and subsequent angles, while others have raised questions about the equations used to describe the motion. There is a mix of interpretations regarding the problem setup and the information provided.

Contextual Notes

Participants note that the problem lacks specific numerical values and that the professor's instructions were not fully conveyed, leading to confusion about the application of formulas and the definitions of variables. There is an emphasis on solving the problem symbolically and understanding the components of motion in both horizontal and vertical directions.

  • #31
jeff12 said:
Okay, so then am I suppose to find the launch angle first and then find the angle at another time after?
No, you have no way to find out the launch angle. It is one of the inputs to the question. Just leave it as an unknown, ##\theta_0##. Rewrite your equation in post #28 using ##\theta_0## and ##v_0## as appropriate... then you are done.

@jeff12, I'm glad you liked my post, but please bring this thread to a satisfactory conclusion by posting that equation using ##\theta_0## and ##v_0## as appropriate.
 
Last edited:
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  • #32
The best way to think about it is to use these ideas:

1. The vertical component of the velocity ##v_y## will be half of its original value. (Can you explain why?)

2. The horizontal component of the velocity ##v_x## stays the same.

3. ##\displaystyle \tan \theta=\frac{v_y}{v_x}##.
 
Last edited:

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