QA 21

#$&*

course PHY 121

Vectors are still confusing!7/25 10

021. projectiles 2

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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?

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Your solution:

vf^2 = 2a`ds

vf^2 = 2 (9.8 m/s^2) (3 m)

vf^2 = 58.8 m^2/s^2

vf = 7.67 m/s

`dt = (vf-v0)/a

`dt = (7.67 m/s - 0 m/s)/9.8 m/s^2

`dt = .8 s

12 m/s * .8 s = 9.6 m

Magnitude: (3m)^2 + (9.6m)^2

9 m^2 + 92.16 m^2

101.16 m^2

magnitude is 10.1 m

angle: tan-1 (3 m/9.6 m) = 17.4 degrees

However, I believe that 17.4 is for the acute angle at the bottom, not off the top where

the projectile is coming from. A triangle has 180 degrees, so the top angle would be

180 - (90 + 17.4) = 72.6 degrees.

confidence rating #$&*:

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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.

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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.

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Self-critique (if necessary):

12 m/s in the x direction and -7.7 m/s in the y direction (since it is going down).

With the Pythagorean Theorum, I ended up with 14.3 m/s, but that's probably close

enough. And for the angle, I got 327 degrees. This makes more sense than everything I

did.

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Self-critique rating: 2

@&

The vector you used was the straight-line vector from the initial to the final point. The path of the ball, however, is curved, starting off less steep than the straight-line path and ending up steeper.

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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?

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Your solution:

vf^2 = 2a`ds

vf^2 = 2 (9.8 m/s^2) (12 m)

vf^2 = 235.2 m^2/s^2

vf = 15.3 m/s

`dt = (vf - v0)/a

`dt = (15.3 m/s - 0 m/s)/9.8 m/s^2

`dt - 1.6 s

confidence rating #$&*:

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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.

Vertical and horizontal motions have different acceleration conditions and must

therefore be analyzed separately and independently. To analyze the vertical motion we

must use the vertical component of the initial velocity. To analyze the horizontal

motion we must use the horizontal component of the initial velocity.

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

its components are

v0_x = 20 m/s * cos(30 deg) and

v0_y = 20 m/s * sin(30 deg).

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.

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Self-critique (if necessary):

I had my original picture correct, and then changed it for an ""over and down"" motion

rather than up and over. I'm still a little stumped on the whole idea of vectors, but

it's starting to sink in.

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Self-critique rating: 2

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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.

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Your solution:

v0=20 m/s

vf = -18.3 m/s

`dt = 2.7 s

`ds/`dt = (v0 + vf)/2

`ds = `dt (vo + vf)/2

`ds = 2.7 s (1.7 m/s) / 2

`ds = 2.295 m

confidence rating #$&*:

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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.

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Self-critique (if necessary):

I need to start thinking in terms of these vectors. I know that vectors have magnitude

and direction and show motion. I'm just having trouble understanding when and how to

use them.

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Self-critique rating: 1

@&

In this case the initial velocity is part vertical, part horizontal. Velocity being a vector quantity, you can use its magnitude and direction to find its horizontal and vertical components.

*@

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.

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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?

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Your solution:

V0y = 12 m/s (sin) (37 deg)

v0y = 7.2 m/s

v0x = 12 m/s (cos) (37 deg)

v0x = 9.6 m/s

The displacement will be +1 m because the projectile is going from 2 m up to 3 m.

vf^2 = v0^2 + 2a`ds

vf^2 = (7.2 m/s)^2 + 2 (-9.8 m/s^2)(1 m)

vf^2 = 51.84 m^2/s^2 + -19.6 m^2/s^2

vf^2 = 32.24 m^2/s^2

vf = 5.7 m/s

vAve = (7.2 m/s + 5.7 m/s)/2

vAve = 6.45 m/s

`dt = 1 m / 6.45 m/s

`dt = .2 s

`ds = vAve * `dt

`ds = 6.45 m/s * .2 s

`ds = 1.29 m

"

@&

.2 seconds is the time required for the projectile to pass the level of the platform on the way up. It has to come back down before striking the platform. It will have traveled 1.29 m in the horizontal direction before doing so.

The key in your analysis is that there are two solutions to vf^2 = 32 m^2/s^2.

vf = +- 5.7 m/s.

Since the projectile has to be descending to land on the platform, you would need to choose the negative value.

This will give you an average velocity of less than 1 meter / second., and will imply a significantly longer time interval, allowing the projectile to complete its arc and land on the platform several meters further along.

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&#Good responses. See my notes and let me know if you have questions. &#