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

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SUMMARY

The discussion focuses on calculating the x- and y-components of a velocity vector represented in a physics problem. The user initially attempted to use the formulas -100 m/s*cos(30 degrees) for the x-component and -100 m/s*sin(30 degrees) for the y-component but received incorrect results. The correct approach involves using cosine for the y-component and sine for the x-component, as clarified by a forum member. This adjustment is crucial for accurately determining the components of the vector.

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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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