Query21

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course Phy 231

7/5/12 @ 11:39 am

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

021. `query 21

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Question: `q Explain how to obtain the final speed and direction of motion of a projectile which starts with known velocity in the horizontal direction and falls a known vertical distance, using the analysis of vertical and horizontal motion and vectors.

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Your solution:

Since the horizontal velocity does not change, it is the same as the initial horizontal velocity. The vertical velocity starts at 0, with acceleration thru a known distance at 9.8 m/s^2 downward. The final vertical velocity is easily found using the fourth equation of motion. So now the Pythagorean Theorem and arctan (vy / vx) the speed and direction of the motion is found.

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Given Solution:

`a** The horizontal velocity is unchanging so the horizontal component is always equal to the known initial horizontal velocity.

The vertical velocity starts at 0, with acceleration thru a known distance at 9.8 m/s^2 downward. The final vertical velocity is easily found using the fourth equation of motion.

We therefore know the x (horizontal) and y (vertical) components of the velocity. Using the Pythagorean Theorem and arctan (vy / vx) we find the speed and direction of the motion. **

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Self-critique (if necessary):

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Question: `qGive at least three examples of vector quantities for which we might wish to find the components from magnitude and direction. Explain the meaning of the magnitude and the direction of each, and explain the meaning of the vector components.

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Your solution:

One would be, a ball that is falling to the ground and has a known angle and velocity. Another might be a car crash. And the third is when there is an x and y component that has the same resultant.

confidence rating #$&*:

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Given Solution:

`a

Examples might include:

A force acting on an object causing it to move in an angular direction.

A ball falling to the ground with a certain velocity and angle.

A two car collision; velocity and momentum are both vector quantities and both important for analyzing the collision..

The magnitude and directiohn of the relsultant is the velocity and direction of travel.

The vector components are the horizontal and vertical components that would produce the same effect as the resultant.

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&#Good responses. Let me know if you have questions. &#