Algebra 2 question

[hea128]

Level: Algebra 2
Question: Imagine a Trial #4 which has an initial speed of 40 m/s. Estimate the maximum height and maximum range of the projectile. Explain your reasoning. Hint: Look for patterns of changes between trials 1, 2, and 3
I have been given this data table to help solve the question along with this simulation website (it does not let you model an Initial Speed of over 30 m/s, but after you input the starting height, angle, and initial speed, it does draw out the parabolas of Trial #1,2,3 and shows which points the parabolas pass through)



So far, I followed the hint and looked for patterns of changes between trials 1,2,3 and found that each trial has a "Range at the Maximum Height" x4 more than the last trial. Using this info, I multiplied 17.66 by 4 yielding 70.64 - so in other words, the x-value of the vertex for Trial #4 is 70.64 meters

This is where I'm stuck... how am I supposed to know the y value of the vertex AND find the roots with only this info? I have not been given any sort of quadratic function for this question.

 

[Deleted member 4993]

Level: Algebra 2
Question: Imagine a Trial #4 which has an initial speed of 40 m/s. Estimate the maximum height and maximum range of the projectile. Explain your reasoning. Hint: Look for patterns of changes between trials 1, 2, and 3
I have been given this data table to help solve the question along with this simulation website (it does not let you model an Initial Speed of over 30 m/s, but after you input the starting height, angle, and initial speed, it does draw out the parabolas of Trial #1,2,3 and shows which points the parabolas pass through)
View attachment 21571


So far, I followed the hint and looked for patterns of changes between trials 1,2,3 and found that each trial has a "Range at the Maximum Height" x4 more than the last trial. Using this info, I multiplied 17.66 by 4 yielding 70.64 - so in other words, the x-value of the vertex for Trial #4 is 70.64 meters

This is where I'm stuck... how am I supposed to know the y value of the vertex AND find the roots with only this info? I have not been given any sort of quadratic function for this question.

For this assignment:

What was the topic of instruction?

Are you supposed to know the physics behind the projectile motion?

 

[hea128]

For this assignment:

What was the topic of instruction?

Are you supposed to know the physics behind the projectile motion?

Here's the thing: No!
We were not taught any sort of physics about this projectile motion. This is the first time I've even heard of it. Topic of instruction would be the Glencoe Algebra 2 Unit 4 which is "Quadratic Functions and Relations"

 

[HallsofIvy]

Level: Algebra 2
Question: Imagine a Trial #4 which has an initial speed of 40 m/s. Estimate the maximum height and maximum range of the projectile. Explain your reasoning. Hint: Look for patterns of changes between trials 1, 2, and 3
I have been given this data table to help solve the question along with this simulation website (it does not let you model an Initial Speed of over 30 m/s, but after you input the starting height, angle, and initial speed, it does draw out the parabolas of Trial #1,2,3 and shows which points the parabolas pass through)
View attachment 21571


So far, I followed the hint and looked for patterns of changes between trials 1,2,3 and found that each trial has a "Range at the Maximum Height" x4 more than the last trial.

that's not quite correct, is it? 4 times 1.1 is 4.4, not 4.41, and 4 times 4.41 is 17.64, not 17,66,

Using this info, I multiplied 17.66 by 4 yielding 70.64 - so in other words, the x-value of the vertex for Trial #4 is 70.64 meters

This is where I'm stuck... how am I supposed to know the y value of the vertex AND find the roots with only this info? I have not been given any sort of quadratic function for this question.

You know that any "quadratic function" is of the form y= ax^2+ bx+ c, with constants a, b, and c, don't you? And putting the given x and y for three data points will give you three equations to solve for a, b, and c.

You have, for range, (5, 4.84), (10, 12.81), and (20, 40.36) so you have
25a+ 5b+ c= 4.84
100a+ 10b+ c= 12.81 and
400a+ 20b+ c= 40.36.

Solve those three equations for a, b, and c, then set x= 40 in ax^2+ bx+ c to find the range for initial speed 40.

 

[JeffM]

First, thank you for the complete problem statement and for telling us what you have thought about. VERY GOOD.

Second, as Subhotosh Khan implied, it is very difficult to give a helpful answer despite your very good initial post because it is not clear what the teaching purpose of the exercise is.

Did you note that the change in range at maximum height is not exactly quadrupling in response to a doubling of initial speed? This frequently occurs with experimental data because measurements are not exact and because minor factors not controlled for have distorting effects. In terms of this problem, are you allowed to ignore this discrepency in doing this problem?

You seem to be aware that a projectile's path can be modeled by part of a parabola and that a quadratic function is graphed by a parabola. Are you allowed to use that knowledge in doing this problem?

 

[hea128]

Second, as Subhotosh Khan implied, it is very difficult to give a helpful answer despite your very good initial post because it is not clear what the teaching purpose of the exercise is.

I'm not entirely sure what the teaching purpose is either, sorry, but I know the unit is "Quadratic Functions and Relations"

Did you note that the change in range at maximum height is not exactly quadrupling in response to a doubling of initial speed? This frequently occurs with experimental data because measurements are not exact and because minor factors not controlled for have distorting effects. In terms of this problem, are you allowed to ignore this discrepency in doing this problem?

Yes, I know that. I presume I'm allowed to ignore this discrepancy because the question asks me to estimate for my answer, so my teacher is probably not looking for an exactly accurate answer.

You seem to be aware that a projectile's path can be modeled by part of a parabola and that a quadratic function is graphed by a parabola. Are you allowed to use that knowledge in doing this problem?

Yes, I think so?

 

[JeffM]

I'm not entirely sure what the teaching purpose is either, sorry, but I know the unit is "Quadratic Functions and Relations"

Given your responses, I am going to guess that the purpose is to teach you how to look for patterns and to see whether you can describe those with a relatively simple mathematical model.

Yes, I know that. I presume I'm allowed to ignore this discrepancy because the question asks me to estimate for my answer, so my teacher is probably not looking for an exactly accurate answer.

I agree that that is probably correct. So let's use 4.

Yes, I think so?

Given the section, that seems reasonable. Thank you for thoughtful answers to my questions.

Halls of Ivy has given you a way to answer this problem that is mathematically correct, but I doubt it is what your teacher expects.

If we model the path of the projectile as a parabola with its axis of symmetry parallel to the y-axis, if we set the origin of our co-ordinate system at ground level directly beneath the launching point, and if we graph our parabola as

[MATH]f(x) = y = ax^2 + bx + c[/MATH],

then what is f(0)?

So, what is c?

 

[hea128]

If we model the path of the projectile as a parabola with its axis of symmetry parallel to the y-axis, if we set the origin of our co-ordinate system at ground level directly beneath the launching point, and if we graph our parabola as

[MATH]f(x) = y = ax^2 + bx + c[/MATH],

then what is f(0)?

So, what is c?

I think c is = 10
That's because my teacher taught me that the c value is the point of the y-intercept, and our y-intercept is 10 so c is 10? I'm not sure ?

But mathematically if I plug in 0 for each x value and solve then I'd also end up with 10

 

[JeffM]

I think c is = 10
That's because my teacher taught me that the c value is the point of the y-intercept, and our y-intercept is 10 so c is 10? I'm not sure ?

But mathematically if I plug in 0 for each x value and solve then I'd also end up with 10

You are doing fine. Yes, each quadratic will have a constant term of 10, given the way we are analyzing the problem.

Notice that I said "each quadratic."

Is there more than one?

There is a different path for each initial speed, right?

So there is a different quadratic for each initial speed. Does that make sense?

Do you know a formula for finding the vertex of a parabola? It probably is somewhere in the section you have been studying.

 

[hea128]

You are doing fine. Yes, each quadratic will have a constant term of 10, given the way we are analyzing the problem.

Notice that I said "each quadratic."

Is there more than one?

There is a different path for each initial speed, right?

So there is a different quadratic for each initial speed. Does that make sense?

Do you know a formula for finding the vertex of a parabola? It probably is somewhere in the section you have been studying.

Yes, there is a different path for each initial speed. Yes I understand each initial speed must have a different quadratic function. Our teacher taught the formula to solve vertex of a parabola is -b over 2a

 

[JeffM]

Great.

Let's define p as

[MATH]p = -\dfrac{b}{2a}[/MATH].

Now we know that p changes as the initial velocity changes. Let's call the initial velocity v.

Can we find an approximate mathematical relationship between v and p?

Remember that YOU found that doubling v resulted in quadrupling p. So the quadratic for when v = 40 = 2 * 20 will have

[MATH]p = 4 * 17.66 = 70.64 \implies 70.64 = - \dfrac{b}{2a} \implies b = WHAT?[/MATH]
Now we have to look for another pattern to give us more information about v, a, and b.

In your work did you find any other pattern?

 

[hea128]

Great.

Let's define p as

[MATH]p = -\dfrac{b}{2a}[/MATH].

Now we know that p changes as the initial velocity changes. Let's call the initial velocity v.

Can we find an approximate mathematical relationship between v and p?

Remember that YOU found that doubling v resulted in quadrupling p. So the quadratic for when v = 40 = 2 * 20 will have

[MATH]p = 4 * 17.66 = 70.64 \implies 70.64 = - \dfrac{b}{2a} \implies b = WHAT?[/MATH]
Now we have to look for another pattern to give us more information about v, a, and b.

In your work did you find any other pattern?

I've looked at this question for hours I really can't figure out what other pattern there is ?

 

[hea128]

Oh wait!!.. I think I found a pattern for the a values
Roughly each trial's a value is x0.25 of the last trial. I am gonna use this information to help me figure out the rest I'll let you know if I have trouble.. thx

 

[Deleted member 4993]

Oh wait!!.. I think I found a pattern for the a values Roughly each trial's a value is x0.25 of the last trial. I am gonna use this information to help me figure out the rest I'll let you know if I have trouble.. thx

Be careful! If you notice - each successive initial speed is doubling from one observation to the next. The range of the projectile is dependent on the initial velocity (among other things).