Calculating Ball Velocity with Unit Vectors

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Homework Help Overview

The discussion revolves around calculating the velocity of a ball shot at a specific speed and angle, focusing on the use of unit vectors to express the velocity components in a two-dimensional context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand how to express the ball's velocity using unit vectors and questions the requirement of the velocity being simply the speed of 5.4 m/s. Some participants explain the process of breaking down the velocity into x and y components using trigonometric functions.

Discussion Status

Participants are exploring the concept of unit vectors and how to represent the velocity vector. There is a suggestion to sketch the vector and find its components, with some guidance on how to project the vector onto the axes. Multiple interpretations of the problem are being discussed, particularly regarding the expression of velocity and its components.

Contextual Notes

There is a mention of potential additional requirements, such as finding the magnitude and direction of the vector with respect to the x-axis, which may not have been fully clarified in the original question.

UrbanXrisis
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If a ball is shot out at 5.4m/s and makes an angle of 30 degrees with the horizontal, the x component would be 5.4cos30 and the y component would be 5.4sin30 correct?

The question then asks me to write an expression for the ball’s velocity, v, using unit vectors for the x-direction and the y-direction. Wouldn’t the velocity be just 5.4m/s? I don’t understand wht they are asking.
 
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Unit vectors are a simple way of describing a 3d (or 2d) vector. Basically, you put everything that describes the x-direction together and put 'ex' with a vector sign over it behind it. Do thesame for the y-direction.
In this case: the x-component of the speed is just 5.4cos(30), so that part of the vector is (5.4cos(30))*ex (actually make that x subscript and add an arrow above the e)

The y-component is thesame thing, only with 5.4sin(30) and ey. In case they ask you to do the expression for t, you'll have to add acceleration like this:
V = (5.4cos(30))*ex + (5.4sin(30) + ..acceleration*t..)*ey
This describes the vector.
 
What if the question wanted me to sketch the vector and find the magnitude and direction of motion with respects to the x-axis?
 
Sorry for the late reply:
Just sketch an arrow making 30 degrees with the horizontal. Then project the vector onto the x-axis and that's your x-component. It's that easy.
Magnitude of the vector in the x-direction is as you described, 5.4cos30. The direction of motion with respect to the x-axis will be 30 degrees.
 

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