Projectile Equation

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mathdad

mathdad

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This is the only question I got wrong in section 4.5. I am currently in 5.1 but would like the set up and solution steps for this projectile equation.


A projectile fired from the point (0, 0) at an angle to the positive x-axis has a trajectory given by

y = cx - (1 + c^2)(g/2)(x/v)^2, where the following must be kept in mind:

x = horizontal distance in meters
y = height in meters
v = initial muzzle velocity in meters per second (m/sec)
g = acceleration due to gravity = 9.81 meters per second squared (m/sec^2)
c > 0 is a constant determined by the angle of elevation.

A howitzer fires an artillery round with a muzzle velocity of 897 m/sec.

A. If the round must clear a hill 200 meters high at a distance of 2000 meters in front of the howitzer, what c values are permitted in the trajectory equation?

B. If the goal in part A is to hit a target on the ground 75 kilometers away, is it possible to do so? If so, for what values of c? If not, what is the maximum distance the round will travel?
 
For part A., we would require:

[MATH]y(2000)>200[/MATH]
The will give us an inequality in the form:

[MATH]ac-b(c^2+1)>200[/MATH]
Is this what you had?
 
MarkFL said:
For part A., we would require:

[MATH]y(2000)>200[/MATH]
The will give us an inequality in the form:

[MATH]ac-b(c^2+1)>200[/MATH]
Is this what you had?
Click to expand...

Yes. I got that far. Can you please, set up Part B? It is the only question that gave me a hard time from section 4.5. Thank you. I will show my work for both parts sometime on Monday.
 
Well, for part B., we require:

[MATH]y(75000)=0[/MATH]
Do you see why?
 
MarkFL said:
For part A., we would require:

[MATH]y(2000)>200[/MATH]
The will give us an inequality in the form:

[MATH]ac-b(c^2+1)>200[/MATH]
Is this what you had?
Click to expand...

The values of a, b, and c are not given. I need to find c values. What is a and b?
 
mathdad said:
I do not see why. You are saying that x = 75,000, right?
Click to expand...

Yes. We are given \(y\) as a function of \(x\), and it is function notation that I'm using. When \(x=75000\), that is when the projectile is 75 km from the point of launch, it will have returned to the ground, where \(y=0\).

mathdad said:
The values of a, b, and c are not given. I need to find c values. What is a and b?
Click to expand...

\(a\) and \(b\) are constants (or parameters), and I was only stating the form you should have, much in the same way that \(2x^2-3x+7\) is of the form:

[MATH]ax^2+b+c[/MATH]
 
MarkFL said:
Yes. We are given \(y\) as a function of \(x\), and it is function notation that I'm using. When \(x=75000\), that is when the projectile is 75 km from the point of launch, it will have returned to the ground, where \(y=0\).



\(a\) and \(b\) are constants (or parameters), and I was only stating the form you should have, much in the same way that \(2x^2-3x+7\) is of the form:

[MATH]ax^2+b+c[/MATH]
Click to expand...

I have something to work with. My work, right or wrong, will be posted.
 
mathdad said:
... c > 0 is a constant determined by the angle of elevation.

A howitzer fires an artillery round ...

... what c values are permitted ...?
Click to expand...
Howitzers have a limited angle of elevation. Did the exercise mention anything about an upper threshhold?

?
 
Otis said:
Howitzers have a limited angle of elevation. Did the exercise mention anything about an upper threshhold?

?
Click to expand...

No. The question did not mention an upper threshold. It would probably mention an upper threshold if stated in a physics book not a college algebra textbook.
 
mathdad said:
... I got that far.
Click to expand...
Were you to have followed the forum's guidelines (show your work), we would have already known that. When you say you got that far, is that the point where you're stuck?


Part B ... is the only question that gave me a hard time ...
Click to expand...
That seems to imply that you had already finished Part A. (I can't tell.) If so, you could have included your answer for that part, so tutors could see where you're stuck. For Part B, did you sketch a diagram? We would prefer to guide you, when helping you set up word problems.

?
 
Otis said:
Were you to have followed the forum's guidelines (show your work), we would have already known that. When you say you got that far, is that the point where you're stuck?



That seems to imply that you had already finished Part A. (I can't tell.) If so, you could have included your answer for that part, so tutors could see where you're stuck. For Part B, did you sketch a diagram? We would prefer to guide you, when helping you set up word problems.

?
Click to expand...

I am working now (on patrol). I will show my full work for this question as I do for other questions on my day off.
 
mathdad said:
... I will show my full work for this question ... on my day off.
Click to expand...
I didn't ask when you would be posting "full work". I was interested in knowing more about what you had already tried.

In the future, please show in the op whatever work you've done on an exercise so far. It's not easy to figure out where to begin tutoring, otherwise. Thank you!

?
 
Otis said:
I didn't ask when you would be posting "full work". I was interested in knowing more about what you had already tried.

In the future, please show in the op whatever work you've done on an exercise so far. It's not easy to figure out where to begin tutoring, otherwise. Thank you!

?
Click to expand...

I always show my work UNLESS there is no starting point for me. In other words, there are questions in the textbook, like most math books, that I simply do not know where to begin. I also make this very clear at the beginning of my post when work is not shown. Ask MarkFL about my math posts. He has been helping me with math for quite a while.
 
MarkFL said:
For part A., we would require:

[MATH]y(2000)>200[/MATH]
The will give us an inequality in the form:

[MATH]ac-b(c^2+1)>200[/MATH]
Is this what you had?
Click to expand...

Part A

0 = 2000c - (1 + c^2)(9.81/2)(2000/987)^2

I got c is approximately 82.0071 and c is approximately 0.0121941.
 
mathdad said:
I always show my work UNLESS there is no starting point for me ...
Click to expand...
Respectfully, no you do not.

You said you missed the question. So obviously you got the wrong answer. Why can't you show what you had tried?

Mark made a guess at what you had tried, and then you confirmed that you had gotten that far. Why didn't you say so up front and show that work?

Please post initial efforts or thoughts in your ops, and please review the forum's submission guidelines.
 
mathdad said:
Part A

0 = 2000c - (1 + c^2)(9.81/2)(2000/987)^2

I got c is approximately 82.0071 and c is approximately 0.0121941.
Click to expand...

What we require is:

[MATH]c(2000)-(1+c^2)\left(\frac{9.81}{2}\right)\left(\frac{2000}{897}\right)^2>200[/MATH]
According to W|A, this implies:

[MATH]\frac{89401-\sqrt{7948807565}}{2180}<c<\frac{89401+\sqrt{7948807565}}{2180}[/MATH]
Decimal approximations:

[MATH]0.11234614383595195986<c<81.906919911209919600[/MATH]
Can you get the exact values algebraically?
 
MarkFL said:
What we require is:

[MATH]c(2000)-(1+c^2)\left(\frac{9.81}{2}\right)\left(\frac{2000}{897}\right)^2>200[/MATH]
According to W|A, this implies:

[MATH]\frac{89401-\sqrt{7948807565}}{2180}<c<\frac{89401+\sqrt{7948807565}}{2180}[/MATH]
Decimal approximations:

[MATH]0.11234614383595195986<c<81.906919911209919600[/MATH]
Can you get the exact values algebraically?
Click to expand...

Mark,

This is my last endeavor to answer parts A and B. My work has been shown over and over again. One last time below. Why did Sullivan include this problem in a college algebra textbook, particularly in a section about quadratic inequalities? Can you tell me why? If my answers for c values are wrong, moving on into Chapter 5, Section 5.1.

Section 5.1 introduces polynomial functions, particularly creating polynomial functions when zeros are given and from graphs of polynomial functions. That should be entertaining enough for the human brain. LOL.

(A) 2000c must exceed
(1 + c^2)(4.905)(2000/897)^2 by at least 200. So, after simplification and pages of wrong answers, I got the following:

(2000c)/23.3845 - 20/23.3845 > 1 + c^2 leading to
82.019c - 8.2019 > 1 + c^2 leading to 0 > c^2 - 82.019c + 9.2019.

The expression on the right will be zero when
c = 41.0095 +/- (1/2)*sqrt(6690.31)
= 41.0095 +/- 40.89715.
So, c must be between 0.10435 and 81.898.

Is this right? If not,I give up!

(B) The maximum range is
(897 m/s)^2/(9.81 m/s^2)
= about 82 km. So, it is possible to hit a target 75 km away. For this, I think I need the following set up:

0 = 75000c - (1+c^2)(4.905)(75000/897)^2.

I went to Wolfram for this baby.

m.wolframalpha.com

Wolfram|Alpha: Making the world’s knowledge computable

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.
m.wolframalpha.com m.wolframalpha.com

Is any of this right? If not, I give up!
 
Let's go back to:

[MATH]c(2000)-(1+c^2)\left(\frac{9.81}{2}\right)\left(\frac{2000}{897}\right)^2>200[/MATH]
Get rid of the decimal value:

[MATH]2000c-(1+c^2)\left(\frac{981}{200}\right)\left(\frac{2000}{897}\right)^2>200[/MATH]
Divide through by 200:

[MATH]10c-(1+c^2)\left(\frac{981}{200}\right)\frac{20000}{897^2}>1[/MATH]
[MATH]10c-(1+c^2)\frac{98100}{897^2}>1[/MATH]
Multiply through by \(897^2=804609\):

[MATH]8046090c-(1+c^2)98100>804609[/MATH]
Arrange in standard form:

[MATH]98100c^2-8046090c+902709<0[/MATH]
Divide through by 9:

[MATH]10900c^2-894010c+100301<0[/MATH]
We know the solution will be the open interval between the roots, which I gave above.
 
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