Finding the y- component of velocity

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Homework Help Overview

The problem involves determining the y-component of velocity for a boat moving at an angle of 60° to the horizontal, given its x-component of velocity as 5 m/s. The context is within the subject area of vector decomposition in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use trigonometric functions to find the y-component, initially applying the cosine function incorrectly. Some participants suggest using a diagram to clarify the relationship between the components and the angle. Others propose using the tangent function as a more appropriate method for this scenario.

Discussion Status

Participants are exploring different trigonometric approaches to solve for the y-component of velocity. There is a mix of suggestions regarding the correct functions to use, and while some guidance has been offered, no consensus has been reached on the best method yet.

Contextual Notes

The original poster is working with limited information, specifically only the angle and the x-component, which may affect the approach to finding the y-component. There is also a mention of available answer choices that do not align with the initial calculations provided.

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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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