QA Assignment 8

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course PHY 121

6/19 4:30

008. Using the Acceleration of Gravity; summarizing the analysis of uniformly accelerated motion

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Question: `q001. There are 5 questions in this set.

The accepted value of the acceleration of gravity is approximately 980 cm/s^2 or 9.8

m/s^2. This will be the acceleration, accurate at most places within 1 cm/s^2, of any

object which falls freely (i.e., without the interference of any other force), near the

surface of the Earth.

If you were to step off of a table and were to fall 1 meter to the ground, without

hitting anything in between, your motion would very nearly approximate that of a freely

falling object.

How fast would you be traveling when you first reached the ground?

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Your solution:

I know: a=9.8m/s^2; `ds=1 m; v0=0

First I figured out 'dt:

v0*`dt + .5a`dt^2 = `ds

First I thought, my initial velocity is 0 so, v0*anything = 0. That made my equation

.5a`dt^2='ds. I multiplied both sides by 2 and got a`dt^2 = 2`ds. Then I divided both

sides by a for `dt^2=2`ds/a. Finally, I took the square root of both sides, so

`dt=`sqrt 2`ds/a. Then I plugged in numbers: `dt = `sqrt 2*1 m/9.8 m/s^2. THis gave

me 2 m / 9.8 m/s^2 or 2 m * 1s^2/9.8 m. I divided 2 m by 9.8 m and got .204 s^2. I

took the square root of .204 s^2 and got .45 s. I took the 9.8 m/s^2 and multiplied by

.45 sec. to find out the change in velocity. This was 4.41 m/s. So the final velocity

would be 4.41 m/s.

confidence rating #$&*:

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Given Solution:

You would have an initial vertical velocity of 0, and would accelerate at 9.8 m/s^2 in

the same direction as your 1 meter vertical displacement.

You would also have a slight horizontal velocity (you don't step off of a table without

moving a bit in the horizontal direction, and you would very likely maintain a small

horizontal velocity as you fell), but this would have no effect on your vertical motion.

So your vertical velocity is a uniform acceleration with v0 = 0, `ds = 1 meter and a =

9.8 m/s^2. The equation vf^2 = v0^2 + 2 a `ds contains the three known variables and

can therefore be used to find the desired final velocity. We obtain

vf = +- `sqrt( v0^2 + 2 a `ds) = +- `sqrt ( 0^2 + 2 * 9.8 m/s^2 * 1 m)= +- `sqrt ( 19.6

m^2 / s^2) = +- 4.4 m/s, approx.

Since the acceleration and displacement were in the direction chosen as positive, we

conclude that the final velocity will be in the same direction and we choose the

solution vf = +4.4 m/s.

STUDENT QUESTION

????how important is the plus or minus before a square root or an approximation of a

square root and what is its significance????

INSTRUCTOR RESPONSE:

In some situations the + solution will be the one that fits the situation, while in

others it will be the -, and in others both the + and - solutions can be valid.

An example:

If a ball is thrown into the air, rising say 20 meters before falling back, then it will

be at the 15 meter height twice, once while rising and once while falling. At one of

those times the velocity will be positive, at the other the velocity will be negative.

So both the + and the - solutions are valid.

On the other hand you might want the velocity when the ball hits the ground. Only for a

downward velocity will be valid for this question. Your equations will give you the +

and the - solution, and you will have to determine which applies (i.e., which is

downward, given your choice of the positive direction).

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Self-critique (if necessary): OK

I did better than I thought. I did not do the +- because the book said that some

negative answers are ""unphysical"" and don't need to be considered.

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Self-critique rating: 3

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Question: `q002. If you jump vertically upward, leaving the ground with a vertical

velocity of 3 m/s, how high will you be at the highest point of your jump?

Note that as soon as you leave the ground, you are under the influence of only the

gravitational force. All the forces that you exerted with your legs and other parts of

your body to attain the 3 m/s velocity have done their work and are no longer acting on

you. All you have to show for it is that 3 m/s velocity. So as soon as you leave the

ground, you begin experiencing an acceleration of 9.8 m/s^2 in the downward direction.

Now again, how high will you be at the highest point of your jump?

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Your solution:

I know that the initial velocity is 3 m/s. The final velocity of this part of the jump

will be 0 m/s. The acceleration is 9.8 m/s.

I used vf = v0 + a`dt. If I'm solving for `dt, I have to make some changes. First I

will subtract v0 from both sides. vf-v0=a`dt. Then I divide both sides by a for (vf-

v0)a = `dt. Then I plugged in the numbers: (0 m/s - 3 m/s)/9.8 m/s^2 = .306 s. To

find the average velocity, I add 0 m/s and 3 m/s then divide by 2, which gives me 1.5

m/s. I then multiply .306 s times 1.5 m/s and I get .459 meters at the height of the

jump.

confidence rating #$&*:

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Given Solution:

From the instant the leave the ground until the instant you reach your highest point,

you have an acceleration of 9.8 m/s^2 in the downward direction.

Since you are jumping upward, and since we can take our choice of whether upward or

downward is the positive direction, we choose the upward direction as positive. You

might have chosen the downward direction, and we will see in a moment how you should

have proceeded after doing so.

For now, using the upward direction as positive, we see that you have an initial

velocity of v0 = + 3 m/s and an acceleration of a = -9.8 m/s^2. In order to use any of

the equations of motion, each of which involves four variables, you should have the

values of three variables. So far you only have two, v0 and a. {}What other variable

might you know? If you think about it, you will notice that when objects tossed in the

air reach their highest point they stop for an instant before falling back down. That

is precisely what will happen to you.

At the highest point your velocity will be 0. Since the highest point is the last point

we are considering, we see that for your motion from the ground to the highest point, vf

= 0. Therefore we are modeling a uniform acceleration situation with

v0 = +3 m/s, a = -9.8 m/s^2 and vf = 0.

We wish to find the displacement `ds. Unfortunately none of the equations of uniformly

accelerated motion contain the four variables v0, a, vf and `ds.

This situation can be easily reasoned out from an understanding of the basic quantities.

We can find the change in velocity to be -3 meters/second; since the acceleration is

equal to the change in velocity divided by the time interval we quickly determine that

the time interval is equal to the change in velocity divided by the acceleration, which

is `dt = -3 m/s / (-9.8 m/s^2) = .3 sec, approx.; then we multiply the .3 second time

interval by the 1.5 m/s average velocity to obtain `ds = .45 meters.

However if we wish to use the equations, we can begin with the equation vf = v0 + a `dt

and solve to find

`dt = (vf - v0) / a = (0 - 3 m/s) / (-9.8 m/s^2) = .3 sec.

We can then use the equation

`ds = (vf + v0) / 2 * `dt = (3 m/s + 0 m/s) / 2 * .3 sec = .45 m.

This solution closely parallels and is completely equivalent to the direct reasoning

process, and shows that and initial velocity of 3 meters/second should carry a jumper to

a vertical height of .45 meters, approximately 18 inches. This is a fairly average

vertical jump.

If the negative direction had been chosen as positive then we would have a = +9.8 m/s^2,

v0 = -3 m/s^2 (v0 is be in the direction opposite the acceleration so if acceleration is

positive then initial velocity is negative) and again vf = 0 m/s (0 m/s is the same

whether going up or down). The steps of the solution will be the same and the same

result will be obtained, except that `ds will be -.45 m--a negative displacement, but

where the positive direction is down. That is we move .45 m in the direction opposite

to positive, meaning we move .45 meters upward.

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Self-critique (if necessary): OK

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Self-critique rating: 3

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Question: `q003. If you roll a ball along a horizontal table so that it rolls off the

edge of the table at a velocity of 3 m/s, the ball will continue traveling in the

horizontal direction without changing its velocity appreciably, and at the same time

will fall to the floor in the same time as it would had it been simply dropped from the

edge of the table.

If the vertical distance from the edge of the table to the floor is .9 meters, then how

far will the ball travel in the horizontal direction as it falls?

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Your solution:

I am not sure about this one, but I'll tell what I'm thinking. The initial velocity

should be 3 m/s. As it falls off the edge of the table, gravity is going to increase

this velocity at the rate of 9.8 m/s^2. From the edge of the table to the floor is .9

m. It it travels .9 m at 3 m/s, it will take .3 s. This does not take any acceleration

into consideration. And this is where I am stuck. I have drawn a picture of a table

with a ball on it. I have marked the path of the ball on the table. At the end of the

table I have written 3 m/s. I have drawn a line that shows .9 m for the height of the

table. I have drawn an arc from the end of the table to the floor. I have written ""How

far?"" at the end of the arc.

confidence rating #$&*:

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Given Solution:

A ball dropped from rest at a height of .9 meters will fall to the ground with a uniform

vertical acceleration of 9.8 m/s^2 downward. Selecting the downward direction as

positive we have

`ds = .9 meters, a = 9.8 m/s^2 and v0 = 0.

Using the equation `ds = v0 `dt + .5 a `dt^2 we see that v0 = 0 simplifies the equation

to `ds = .5 a `dt^2, so

`dt = `sqrt( 2 `ds / a) = `sqrt(2 * .9 m / (9.8 m/s^2) ) = .42 sec, approx..

Since the ball rolls off the edge of the table with only a horizontal velocity, its

initial vertical velocity is still zero and it still falls to the floor in .42 seconds.

Since its horizontal velocity remains at 3 m/s, it travels through a displacement of 3

m/s * .42 sec = 1.26 meters in this time.

STUDENT COMMENT:

I don’t see why you don’t use the formula that I used in my answer.

INSTRUCTOR RESPONSE:

In your solution you chose to solve the third equation

ds = v0 * dt + 0.5 * a * dt^2

for `dt. Since v0 = 0 for the vertical motion, the equation simplifies and is easy to

solve for `dt.

However I avoid using the third equation to solve for `dt because it is quadratic in

`dt, and is therefore very confusing to most students. It is less confusing to use the

fourth equation to find vf, after which we can easily reason out `dt.

Of course in the present case v0 = 0 and the equation becomes every bit as simple as the

fourth equation; in fact when v0 = 0 it's simpler to use the equation you used. However

most students have problems with special conditions and special cases, so I choose not

to confuse the issue, and consistently use the fourth equation in my solutions.

My convention of using the fourth equation in this situation does no harm to students

who understand how to solve the quadratic, and who know how consider and apply special

conditions. You and other students who are sufficiently comfortable with the mathematics

should always use the most appropriate option.

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Self-critique (if necessary):

That was the formula I had written down. I was sitting here wishing that the initial

velocity was zero so that it would be easier to solve. I did not interpret that having

it roll off the table at 3 m/s would yield the same result as dropping it from zero m/s.

I was on the right path, just misinterpreted the initial velocity.

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Self-critique rating: 2

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 ball starts out with a horizontal velocity of 4 m/s and vertical

velocity 0 m/s. It falls freely for 2 seconds.

During this 2 seconds how far does it travel in the vertical direction, and how far in

the horizontal direction?

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Your solution:

I am going to need more work on this. I have done the following:

If a ball goes across a horizontal table at the velocity of 4 m/s for 2 s, it will be

travel 8 m.

If a ball drops from 0 m/s and is accelerating because of the pull of gravity at the

rate of 9.8 m/s^2, for 2 seconds it will attain the velocity of 19.6 m/s. If you take

the initial velocity of 0 m/s and add it to 19.6 m/s then divide by 2 for the average,

you will get 9.8 m/s. If you multipy by 2 for the amount of time, you will get 19.6 m.

However, I'm reading this as a total of 2 seconds, and I'm not sure how to split these

amounts up into a horizontal amount and a vertical amount.

If I split it in half, then it would go 4 m horizontally and 9.8 m vertically.

confidence rating #$&*:

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Question: `q005. A ball starts out with a horizontal velocity of 4 m/s and vertical

velocity 0 m/s. It falls freely to the ground 20 meters below.

During its fall how far does it travel in the vertical direction, and how far in the

horizontal direction?

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Your solution:

Known: for the vertical drop-acceleration is 9.8 m/s^2, v0=0 m/s; `ds = 20 m

Using `ds=v0`dt + .5a`dt^2 and solving for `dt, remembering that v0=0 m/s, you can

eliminate the v0`dt because any number times zero always equals zero. That leaves you

with: .5a`dt^2 = `ds. You can multiply both sides by 2 to move the .5 and divide by a

to move the acceleration. This gives you `dt^2 = 2`ds/a. Then you have to take the

square root of both sides to get the answer for `dt, so the 2`ds/a part will be

(2*20m)/9.8 m/s^2, or 40 m * 1 s^2/9.8 m, which gives you 4.08 s^2. The next step is to

take the square root of 4.08, which is approximately 2 seconds.

For the horizontal movement, you know that it is 4 m/s and that it will be dropping for

2 s. This will give you a horizontal displacement of 8 m.

confidence rating #$&*:

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Self-critique (if necessary):

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Self-critique rating: