Finding the y- component of velocity

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SUMMARY

The discussion focuses on calculating the y-component of velocity for a boat moving at a 60° angle to the horizontal, with a known x-component of 5 m/s. The correct approach involves using the tangent function, specifically the equation y = 5 * Tan(60°), which yields a y-component of 5√3 m/s. Participants clarified the use of trigonometric functions, emphasizing the importance of identifying the correct operation for finding the opposite side of the triangle formed by the velocity components.

PREREQUISITES
  • Understanding of basic trigonometry, specifically SOH CAH TOA.
  • Familiarity with vector components in physics.
  • Knowledge of how to apply the tangent function in calculations.
  • Ability to interpret and draw diagrams representing angles and vectors.
NEXT STEPS
  • Learn how to apply trigonometric functions in physics problems.
  • Study vector decomposition in two-dimensional motion.
  • Explore the use of sine and cosine functions for different angle scenarios.
  • Practice solving problems involving angles and components using diagrams.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and vector analysis, as well as educators seeking to enhance their teaching of trigonometric applications in real-world scenarios.

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Homework Statement



A boat moves at an angle of 60° to a horizontal line. The x component of the boat's velocity is 5 m/s. What is its y component of velocity?
Question 25 answers

10 m/s

10 √3 m/s

5 √3 m/s

5 m/s



Homework Equations



How exactly do I find the y-component for this problem?

The Attempt at a Solution


I originally drew a diagram of a triangle and came up with the equation cos 60=x/5, when I tried to solve for x I got 2.5 m/s. This isn't one of the solutions available so I am thinking I'm not using the correct equation.
 
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cos 60 should be equal to the vector adjacent to the 60 degree angle divided by the hypotenuse. You're looking for the vector opposite from the angle, so there's a different operation for that.
 
Flipmeister said:
cos 60 should be equal to the vector adjacent to the 60 degree angle divided by the hypotenuse. You're looking for the vector opposite from the angle, so there's a different operation for that.

i would start out with a diagram, showing the angle and x and y velocities.

it makes the trig much easier
 
Hi,
as your not given the hypotenuse your are only given an angle and the x component you need to use the tangent function

SOH CAH TOA

This then becomes 5 * Tan(60) = the y component you are after.

71vk0.jpg


I hope that helps and more importantly I hope I am right
thanks
Mr C
 

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