Projectile motion over an inclined plane

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SUMMARY

The discussion centers on the analysis of projectile motion over an inclined plane, specifically addressing the calculations involving the vertical and horizontal components of motion when launching a projectile at 20 m/s perpendicular to a 30° slope. The participant questions the removal of the term "– 1/2 × 10 × t^2" from the vertical component equation when equating it to the tangent of 30°. It is clarified that this term is not removed but is implicitly accounted for in the overall motion analysis. The relevant equations utilized include the Suvat equations of motion.

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  • Understanding of projectile motion principles
  • Familiarity with Suvat equations of motion
  • Basic trigonometry, particularly tangent functions
  • Knowledge of inclined plane dynamics
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MattDutra123
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Homework Statement
A projectile is launched perpendicular to a 30° slope at 20 m/s. Calculate the distance between the launching position and landing position.
Relevant Equations
Suvat equations of motion.
I have attached below the solution provided by the textbook. I understand the whole solution, except when they divide the horizontal component and the vertical component and equate it to the tangent of 30. What seems strange to me is that, while initially the equation for the vertical component was: "y = 20 × cos30° × t – 1/2 × 10 × t^2", when they equate it to the tangent they remove the "– 1/2 × 10 × t^2" section entirely. Is this a careless mistake by the book, or is there another reason for the removal of that section of the equation?
Thanks in advance.
 

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MattDutra123 said:
Problem Statement: A projectile is launched perpendicular to a 30° slope at 20 m/s. Calculate the distance between the launching position and landing position.
Relevant Equations: Suvat equations of motion.

I have attached below the solution provided by the textbook. I understand the whole solution, except when they divide the horizontal component and the vertical component and equate it to the tangent of 30. What seems strange to me is that, while initially the equation for the vertical component was: "y = 20 × cos30° × t – 1/2 × 10 × t^2", when they equate it to the tangent they remove the "– 1/2 × 10 × t^2" section entirely. Is this a careless mistake by the book, or is there another reason for the removal of that section of the equation?
Thanks in advance.
1/2 × 10 × t^2 is not removed.
 

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