Query 21

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course Phy 121

2

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:

 

 vf can easily be found using the know quantities of the vertical motion. Using v0=0m/s, the given quantity for ds, and a=9.8m/s^2, vf can be solved for using the equations of uniformly accelerated motion.

The direction of the vector can be found by solving for the final velocity in the vertical direction and then using the formula: arctan(velocity_y/velocity_x)

 

 

confidence rating #$&*:

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3

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Given Solution:

ok

`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):

ok

 

 

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Self-critique Rating:

ok

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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:

two objects colliding: the magnitude might be the speed or force and the direction would be the angle of impact, using the vectors you could determine where the two objects would travel after the collision.

An object traveling in one direction and then, another force acts on it from a different angle. The second vector acting in the different direction would cause the objects original trajectory to change.

Two forces putting tension on an object in opposite direction, an analysis of the magnitudes of each vector would determine in which direction it would travel.

 

 

confidence rating #$&*:

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3

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Given Solution:

ok

`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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Self-critique (if necessary):

 ok

 

 

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Self-critique Rating:

ok

 

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Self-critique (if necessary):

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Self-critique rating:

ok

 

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Self-critique (if necessary):

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Self-critique rating:

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&#Very good responses. Let me know if you have questions. &#