What are the x- and y-components of the velocity vector?

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The discussion focuses on determining the x- and y-components of a velocity vector from a provided figure. The initial calculations for the components were incorrect due to the misuse of sine and cosine functions. The correct approach involves recognizing that the cosine function should be applied to the component along the negative y-axis, while the sine function applies to the x-axis. A user sought clarification on the angles formed with the axes after correcting their understanding. The conversation highlights the importance of accurately interpreting vector components in physics problems.
jamal.kun
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Homework Statement


Question is: What are the x- and y-components of the velocity vector shown in the figure?
knight_Figure_03_22.jpg


https://session.masteringphysics.com/problemAsset/1384003/6/knight_Figure_03_22.jpg

The Attempt at a Solution


I tried -100 m/s*cos(30 degrees)=-86.6m/s for the x component and -100m/s*sin(30 degrees)=-50m/s for the y component, but was incorrect for both. What am I missing?
 
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jamal.kun said:

Homework Statement


Question is: What are the x- and y-components of the velocity vector shown in the figure?
knight_Figure_03_22.jpg


https://session.masteringphysics.com/problemAsset/1384003/6/knight_Figure_03_22.jpg

The Attempt at a Solution


I tried -100 m/s*cos(30 degrees)=-86.6m/s for the x component and -100m/s*sin(30 degrees)=-50m/s for the y component, but was incorrect for both. What am I missing?

Welcome to the PF.

You got the sin & cos backwards. Try drawing the line from the tip of the vector to the negative y axis. See how the component of the vector on the negative y-axis involves the cos(30 degrees)?
 
Thanks berkeman, I see what you mean now. I didn't expect something like that. Thank you for the incredibly quick reply.
 
If we drawing the line from the tip of the vertor to the negative y-axis then What will be its angle with x axis and negative y-axis?
 
Abdullah Wali said:
If we drawing the line from the tip of the vertor to the negative y-axis then What will be its angle with x axis and negative y-axis?
If you are asking jamal.kun, the thread is seven years old and jamal is no longer a forum member.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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