In a Projectile Motion Question, How to Find Theta?

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SUMMARY

The discussion centers on calculating the angle theta in a projectile motion problem using the formula tan-1(vertical velocity/horizontal velocity). The specific calculation provided is tan-1(30/7), resulting in an angle of approximately 77 degrees. The participants confirm that the method of using the inverse tangent function to determine the angle from the velocity components is accurate. This establishes a clear understanding of how to derive the angle in projectile motion scenarios.

PREREQUISITES
  • Understanding of basic physics concepts, specifically projectile motion.
  • Familiarity with vector components, particularly vertical and horizontal velocities.
  • Knowledge of trigonometric functions, especially the inverse tangent function.
  • Ability to perform calculations involving angles and ratios.
NEXT STEPS
  • Study the principles of projectile motion in greater depth.
  • Learn how to apply trigonometric functions in physics problems.
  • Explore vector decomposition in two-dimensional motion.
  • Practice solving various projectile motion problems using different initial velocities.
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in mastering the concepts of projectile motion and vector analysis.

Sebastian_
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Homework Statement
Excuse me for asking such a simple question, but, in a projectile motions question, if given an initial horizontal speed of 7 mps and a initial vertical speed of 30 mps how do I find theta? (the angle at launch)
Relevant Equations
I was told this was the equation for theta so: tan-1(Voy/Vox)
Solving the relevant equation tan-1(30/7) = 76.865977693604 = 77

Just need confirmation or a correction since I am very new to physics in general
 
Physics news on Phys.org
Velocity is a vector with magnitude and direction. The vertical and the horizontal velocity are its components in x and y direction. To find out its angle relative to the horizontal, just divide the vertical with the horizontal and take the inverse tangent of it. Your answer is correct.
 
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Thank you very much for clearing that up
 

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