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.  projectiles 2

 

 

Question:  `q001.  Note that this assignment contains 4 questions.

 

.  A projectile has an initial velocity of 12 meters/second, entirely in the  horizontal direction.  At the instant of first contact with a level floor three meters lower than the initial position, what will be the magnitude and direction of the projectile's  velocity vector?

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating:
 

Given Solution: 

At the instant of first contact, any force between the ball and the floor will not have had time to affect the motion of the ball.  So we can assume uniform acceleration throughout the interval from the initial instant to the instant of first contact.

 

To answer this question we must first determine the horizontal and vertical velocities of the projectile at the  instant it first encounters the floor.  The horizontal velocity will remain at 12 meters/second.  The vertical velocity will be the velocity attained by a falling object which is released from rest and allowed to fall three meters under the influence of  gravity.

 

Thus the vertical motion will be characterized by initial velocity v0 = 0, displacement `ds = 3 meters and acceleration a = 9.8  meters/second ^ 2.  The fourth equation of motion, vf^2 = v0^2 + 2 a `ds, yields

 

final vel in y  direction:  vf = +-`sqrt( 0^2  + 2 * 9.8 meters/second ^ 2 * 3 meters) = +-7.7 meters/second.  Since we took the acceleration to be in the positive direction the final velocity will be + 7.7 meters/second.

 

This final velocity is in the downward  direction.  On a standard x-y coordinate system, this velocity will be directed along the negative y axis and the final velocity will have y coordinate -7.7 m/s  and x coordinate 12 meters/second.

 

The magnitude of the final velocity is therefore `sqrt((12  meters/second) ^ 2 + (-7.7 meters/second) ^ 2 ) = 14.2 meters/second, approximately.

 

The direction of the final velocity will therefore  be arctan ( (-7.7 meters/second) / (12 meters/second) ) = -35 degrees, very approximately, as measured in the counterclockwise direction  from the positive x axis.  The direction of the projectile at this instant is therefore 35 degrees below horizontal. This angle is more commonly expressed as 360 degrees + (-35 degrees) = 325 degrees.

 

STUDENT QUESTION

 

Why did the y component of 7.7m/s become negative?

INSTRUCTOR RESPONSE

 

When we solved for the time of fall we assumed the downward direction to be positive.

When we put the entire picture on an xy coordinate plane we had to change our choice of positive direction to agree with
the orientation of the y axis.

 

STUDENT QUESTION

 

 I’m still confused by this part of the equation vf = +-`sqrt( 0^2 + 2 * 9.8 meters/second ^ 2 * 3 meters) = +-7.7
meters/second. I don’t understand where the +- part comes from.

INSTRUCTOR RESPONSE

 

The solution to any equation of the form x^2 = c is x = +- sqrt(c).

For example if x^2 = 25, then x = sqrt(25) = 5 is a solution, but so is x = -sqrt(25) = -5.
Another way of writing these solutions is 25^(1/2) = 5 and -(25^(1/2)) = - 5.

It is similar with any even power. For example x^4 = c has two solutions, x = c^(1/4) and x = - c^(1/4).

 

STUDENT QUESTION:

 

Hmm. I see where I made a mistake, but I’m not too sure why I made the mistake.


I don’t get why the vertical velocity is negative. I had already established the direction of acceleration due to gravity
as the positive direction, so I figured the vertical velocity would be positive as well.

 

Question: is the vertical velocity negative because in relation to an x-y plane, the downward motion of the ball relates to the negative aspect of the y-axis.

INSTRUCTOR RESPONSE: As with any vector, the angle of the final velocity vector is to be specified using a standard x-y coordinate system.

The downward direction was chosen as positive when finding the vertical velocity. The conclusion of this analysis was that the final velocity is 7.7 m/s downward. The standard x-y coordinate system was not used in this part of the solution, and is not necessary at this point.

However to specify the direction of the final velocity vector, we must now place the horizontal and vertical velocities on a standard x-y system, with the x axis to the right and the y axis upward. On this coordinate system the y velocity, being downward, has to be represented by a vector whose vertical coordinate is negative, as in the given solution.

 

STUDENT COMMENT

 

 I must have punched something into the calculator incorrectly because I didn’t end up getting -35 for the direction of the
final velocity. If I had calculated this correctly and added 360 to it then it would have given me 325 degrees.

INSTRUCTOR RESPONSE

 

You got an angle of less than 1 degree, which wouldn't make sense in terms of the quantities you used.

 

You probably had your calculator in radian mode. To get the angle in degrees the calculator has to be in degree mode.

Be sure you always sketch the situation. If the result you get from the calculator is inconsistent with your sketch, as would be the case here, you need to double-check your calculations. If the inconsistency involves an angle you calculated, then the first thing to check is the mode.

 

STUDENT COMMENT

 

Ok, I think I understand, but it is still a little blurry.
It seems simple once you figure everything out, but getting everything in the correct place is my problem.

INSTRUCTOR RESPONSE

 

You can solve the vertical velocity and the horizontal velocity for this interval.

The velocity at impact will therefore be a vector with known vertical and horizontal components.

 

STUDENT COMMENT and QUESTION

 

i got -33 degrees not -35 an i'm not sure why or if it matters. also can you explain a little more of why we have to add 360
to this number? i didn't do that so would that make this value incorrect?
 
We generally want the angle of a vector to be between 0 and 360 degrees, if we're working in degrees, or between 0 and 2 pi if we're working in radians. It saves us having to think about things like positive and negative, and from having to add or subtract multiples of 360 degrees.

The estimated arctan used in the given solution is only an approximation. If you calculated the actual arctangent and got -35 degrees, chances are very good that your result is correct.

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:
 

Question:  `q002.  A projectile is given an initial velocity of 20 meters/second at an angle of 30 degrees above horizontal, at an altitude of 12 meters above a level  surface.  How long does it take for the projectile to reach the level surface? 

 

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating:
 

Given Solution: 

To determine the time required to reach the level surface we need only analyze the vertical motion of the  projectile.  The acceleration in the vertical direction will be 9.8 meters/second ^ 2 in the downward direction, and the displacement will be 12 meters in the downward direction.  Taking the initial velocity to be upward and to the right, we situate our x-y coordinate  system with the y direction vertically upward and the x direction toward the right.  Thus the initial velocity in the vertical direction will be equal to the y component of the initial velocity.  Assuming the horizontal direction to be to the right, the angle of the velocity vector with the positive x axis is 30 degrees, so

 

v0y = 20 meters/second *  sine (30 degrees) = 10 meters/second.

 

Characterizing the vertical motion by v0 = 10 meters/second, `ds = -12 meters (`ds is downward  while the initial velocity is upward, so a positive initial velocity implies a negative displacement), and a = -9.8 meters/second ^ 2, we see  that we can find the time `dt required to reach the level surface using either the third equation of motion `ds = v0 `dt + .5 a `dt^2, or we can use the fourth equation vf^2 = v0^2 + 2 a `ds to find vf after which we can easily find `dt.  To avoid  having to solve a quadratic in `dt we choose to start with the fourth equation.

 

We obtain vf = +-`sqrt ( (10 meters/second) ^ 2 + 2 *  (-9.8 meters/second ^ 2) * (-12 meters) ) = +-18.3 meters/second, approximately.  Since we know that the final velocity  will be in the downward direction, we choose vf = -18.3 meters/second.

 

We can now find the average velocity in the y direction.   Averaging the initial 10 meters/second with the final -18.3 meters/second, we see that the average vertical velocity is -4.2  meters/second.  Thus the time required for the -12 meters displacement is `dt = `ds / vAve = -12 meters/(-4.2 meters/second) = 2.7 seconds.

 

STUDENT QUESTIONS:

 

 Why did we use   v0y = 20 meters/second * sine (30 degrees) = 10 meters/second?

 

Why would there be an initial velocity?

 

I don’t understand why we would use sin instead of tan.
 

INSTRUCTOR RESPONSE:

 

A condition of the problem is that the initial velocity is 20 m/s at angle 30 degrees. The initial velocity is therefore a vector with vertical and horizontal components.

The vector v0 has magnitude 20 m/s and makes angle 30 deg with the positive x axis, so its components are

If you are thinking in terms of triangles, the 20 m/s quantity is the hypotenuse of the triangle and the components are the legs of the triangle, which are found using the sine and cosine.

 

For reference, the first figure below depicts the right triangle defined by the vector; the second includes the components (the components are the green vectors along the coordinate axis).

 

    

STUDENT QUESTION

 

 Why is the v0 not equal to 0m/s. I thought with ALL projectiles, it is
inferred that the initial velocity in the vertical direction is 0 m/s. And what I have gotten from this explanation, what we
have previously learned as the altitude to represent the height of the table, now we have to basically look at altitude as
representing the horixontal distance…. in terms of the way the vector triangle is twisted in its representation.

INSTRUCTOR RESPONSE

 

In the problems we've seen to this point, the initial vertical velocity was zero. The reason is that the analysis is often much easier when the initial vertical velocity is zero. We learn the principles of the analysis using the simplest possible system that illustrates the important features of the process; then we move beyond the simplest cases and apply those principles to slightly more complicated conditions.

The projectile in this situation was given an initial velocity of 20 meters/second at an angle of 30 degrees above horizontal. An object moving at 30 degrees to horizontal has a nonzero vertical velocity, which we find using the principles of vector components.

 

In this problem the only difference with earlier projectile problems is that you have to include a nonzero vertical velocity when finding the time of fall and the final vertical velocity.

 

STUDENT QUESTION

 

To solve for the problem, I should have taken 20 m/s * sin (30 degrees). This equals 10 m/s. What value do we call this
though exactly?  Is it the vertical motion?

INSTRUCTOR RESPONSE  20 m/s is the initial velocity of the object, generally designated v0.

10 m/s is the vertical component of the initial velocity, or the initial velocity in the vertical direction.  This is generally designated v0_y, as it was in the given solution.

 

STUDENT QUESTION

 

Shouldn't the average velocity be about 14.2 m/s (( 10 m/s + 18.3 m/s )/2 ) and not 4.2 m/s, making the time it takes about
.85 seconds???

INSTRUCTOR RESPONSE

 

The initial vertical velocity is upward, the final vertical velocity is downward. So the two velocities have different signs. When added and divided by 2, we get an average velocity of magnitude 4.2 m/s.

If the projectile started at angle 30 deg below horizontal the average velocity would have magnitude 14.2 m/s and the fall would take only .85 sec. Starting above horizontal, though, the ball first rises to a height greater than the 12 m, which takes time, then falls back down.

 

STUDENT QUESTIONS

 

how can you assume that the initial velocity is upward into the right? is this because it's angle is 30 degrees and if so why
couldn't it be tilted in the other direction, i.e. downward and to the left?

 

is the average velocity of -4.2 m/s only the
velocity in the vertical direction still?


i'm sorry but these problems really throw me off!
 
INSTRUCTOR RESPONSE 

 

In any problem we have to specify our coordinate system and the direction chosen for the given vector quantities.

 

All we know is that the initial velocity is at 30 degrees above horizontal.  The horizontal direction goes both right and left.  So it would have been perfectly valid to assume the initial velocity to be up and to the left.  However we have to make one assumption or the other.  The choice made in this problem results in fewer negative signs, so is probably the more natural choice.

 

We have implicitly chosen our coordinate system to be the 'standard' x-y system with x to the right and y upward. Had we chosen to have the initial velocity to be upward and to the left the initial velocity would have been at 150 degrees, as
measured counterclockwise from the positive x axis.

 

The choice in the given solution was stated in the phrase 'Taking the initial velocity to be upward and to the right', so that our initial direction is into the first quadrant, at angle 30 degrees. The later phrase 'Assuming the horizontal direction to be to the right' reiterates this choice.


Starting with the phrase 'Characterizing the vertical motion by v0 = 10 meters/second, `ds = -12 meters (`ds is downward 
while the initial velocity is upward, so a positive initial velocity implies a negative displacement), and a = -9.8
meters/second ^ 2', the rest of the given solution proceeds to solve for `dt, using these quantities. So the remainder of
the given solution is for the vertical motion. The -4.2 m/s avera ge velocity therefore applies to the vertical motion.

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:
 

Question:  `q003.  What will be the horizontal distance traveled by the projectile in the preceding exercise, from the initial instant to the instant the projectile strikes  the flat surface.

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating:
 

Given Solution: 

The horizontal velocity of the projectile will not change so if we can find this horizontal velocity, knowing  that the projectile travels for 2.7 seconds we can easily find the horizontal range.

 

The horizontal velocity of the projectile is simply  the x component of the velocity:

 

horizontal velocity = 20 meters/second * cosine (30 degrees) =  17.3 meters/second.

 

Moving at this rate for 2.7 seconds the projectile travels distance 17.3 meters/second * 2.7 seconds = 46 meters,  approximately.

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

 

 

 

If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily.  If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.

 


 

Question:  `q004.  A projectile is given an initial velocity of 12 meters / second in a direction 37 degrees above horizontal.  It is released from a height of 2 meters and lands on a platform whose height is 3 meters.  How much time elapses, and how far does it travel in the horizontal direction between release and landing?

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating:

Self-critique (if necessary):

 

Self-critique rating: