another systems of equations problem

[allegansveritatem]

Here is exercise:

I can't see my way to a solution for part a) (let alone part b) ). What I have done is draw a diagram and list all the things I know about the situation. (see below) Can anyone suggest what to do next?

 

[Steven G]

What is the acceleration in the x direction, a_x? How about the a_y?
What is the velocity in the x direction, v_x and v_y?
What is the displacements s_x and s_y?
What is the (x,y) of where the ball lands and where the ball is launched? Note that the answer will depend of where you place the origin.

 

[Steven G]

You really should know the answer to my last question! You have a right triangle where the two legs are in the ratio of 3 to 4. If they were 3 and 4, then the hypotenuse would be 5. But the hyp is 50, 10 times 5. So the other two sides are 30 and 40. You know this!

 

[allegansveritatem]

You really should know the answer to my last question! You have a right triangle where the two legs are in the ratio of 3 to 4. If they were 3 and 4, then the hypotenuse would be 5. But the hyp is 50, 10 times 5. So the other two sides are 30 and 40. You know this!

yes, I can find the dimensions of the two triangles but how is that going to get me to the vertex of the parabola? I vaguely recall something about a vertex form of a quadratic equation and I will have to dig out one of my algebra books and review that. It may even be in the text I'm using now but about 400 pages back which means I haven't looked at that for well over a year. I started studying this book in March of 2019 and have not missed putting in 2 hours a day on it until now and I don't expect to finish it for probably another year and maybe not even then.

 

[Deleted member 4993]

Here is exercise:View attachment 21289View attachment 21290

I can't see my way to a solution for part a) (let alone part b) ). What I have done is draw a diagram and list all the things I know about the situation. (see below) Can anyone suggest what to do next?
View attachment 21291

The trajectory of the ball is:

y = ax^2 + bx [ x=0 has to be one of the roots since we are starting from the origin)

The equation for the elevation-line of the hill will be y = -(3/4)x

If the ball lands at a point (x₁, y₁), calculate x₁ and y₁ [or you can follow Jomo's intuitive answer]

 

[allegansveritatem]

Thinking about this today I had a thought: Seems to me that the parabola could very well be approximated by a right triangle with angles 45, and 45. Can anyone see this?One leg of the triangle would be from the origin to the vertex and the other from the vertex to the x axis. As for the suggestion of Jomo, I don't see when he is tending with it. How does knowing the dimensions of the right triangle under the x axis help us find the vertex? I will have to study your post to see if I can make a connection between it and the problem and Jomo's post. I'll be back. (it suddenly occurs to me that I am forgetting that this is to be solved using the systems of equation method so not taking that into consideration may be why I haven't seen what you and Jomo are pointing to.)

 

[Steven G]

It has been forever since I tried to derive the formulas below. They look fine to be(!)
a_x=0
v_x = v₀cos(theta)
s_x = v₀cos(theta)t + s_0x

a_y = -g
v_y = -gt + v₀sin(theta)
s_y = -gt^2/2 + v₀sin(theta)t + s_0y

This will help you get your quadratic equation.
Note that to get the above formulas all I did was integrate a_y = -g twice and a_x= 0 twice.

 

[Steven G]

The trajectory of the ball is:

y = ax^2 + bx [ x=0 has to be one of the roots since we are starting from the origin)

The equation for the elevation-line of the hill will be y = -(3/4)x

If the ball lands at a point (x₁, y₁), calculate x₁ and y₁ [or you can follow Jomo's intuitive answer]

Subhotosh, I disagree with you when you say that the starting point is the origin. Although I can't think of one now there has to be examples where placing the starting at the origin is not the best choice. I know that you are a great instructor and helper (seriously!) but let the OP figure out that they get to place the origin where ever they want.

 

[Deleted member 4993]

Subhotosh, I disagree with you when you say that the starting point is the origin. Although I can't think of one now there has to be examples where placing the starting at the origin is not the best choice. I know that you are a great instructor and helper (seriously!) but let the OP figure out that they get to place the origin where ever they want.

In the picture (response #1) - the given plot of the trajectory started at point of intersection of the x and y axes (origin).

 

[allegansveritatem]

It has been forever since I tried to derive the formulas below. They look fine to be(!)
a_x=0
v_x = v₀cos(theta)
s_x = v₀cos(theta)t + s_0x

a_y = -g
v_y = -gt + v₀sin(theta)
s_y = -gt^2/2 + v₀sin(theta)t + s_0y

This will help you get your quadratic equation.
Note that to get the above formulas all I did was integrate a_y = -g twice and a_x= 0 twice.

you have soared far above and way beyond me with this. I have no idea what g or s refer to. Please, give me the "for dummies" version of this.

 

[allegansveritatem]

The trajectory of the ball is:

y = ax^2 + bx [ x=0 has to be one of the roots since we are starting from the origin)

The equation for the elevation-line of the hill will be y = -(3/4)x

If the ball lands at a point (x₁, y₁), calculate x₁ y₁ [or you can follow Jomo's intuitive answer]

very early on with this problem I did find the equations for the lines, ie, y=x and y=-3/4 x. From this I found that one solution could only be zero but I did not think that was getting me where I wanted to go. I didn't know how to use the information. Well, I will go at this again today armed with the advice in this thread and with the high hope that my ignorance is vincible or at least subject to reduction.

 

[HallsofIvy]

"g" is standard notation for the downward acceleration due to gravity and "sx" and "sy" are standard notation for x and y coordinates of position. However, this problem is giving you a different notation so you don't want to use those. In this problem the position of the ball is given by x and y and the acceleration is incorporated into "a".

In this problem, you are told that "using Calculus it can be shown that the path of the ball is given by y= ax^2+ x+ c for some constants a and c". So I take it you have not yet taken Calculus and are not expected to derive that. Since "real life" problems do not have a coordinate system attached you do need to select one. It is standard practice to have the x-axis horizontal and the y-axis vertical and the equation "y= ax^2+ x+ c" is assuming that. You do have to decide where to put the origin of the coordinate system. I see two points where it would be reasonable to set the origin- at the initial point of the trajectory or and the end point. Choosing that initial point as the origin, x= 0 and y= 0 we must have y= 0= a(0^2)+ b(0)+c= c. So now we know the equation is y= ax^2+ x. The end point is 50 m down the hill which has a "slope of -3/4". The slope (not quite Calculus) is "y/x= -3/4" so y= -3x/4. Using the Pythagorean formula, x^2+ y^2= x^2+(3/4)^2x^2=(1+9/16)x^2= (25/16)x^2= 50^2 so x^2= (50^2)(16/25)= (2)(50)(16)=(4)(25)(16). Taking the square root of both sides, x= (2)(5)(4)= 40. y=-3x/4= -30. In coordinate system with origin at the starting point the end point has coordinates x= 40, y= -30. Putting those into y= ax^2+ x, -30=a(40)^2+ 40. -70= 1600a so a= -7/160.

 

[jonah2.0]

Beer soaked hint follows.

"g" is standard notation for the downward acceleration due to gravity and "sx" and "sy" are standard notation for x and y coordinates of position. However, this problem is giving you a different notation so you don't want to use those. In this problem the position of the ball is given by x and y and the acceleration is incorporated into "a".

In this problem, you are told that "using Calculus it can be shown that the path of the ball is given by y= ax^2+ x+ c for some constants a and c". So I take it you have not yet taken Calculus and are not expected to derive that. Since "real life" problems do not have a coordinate system attached you do need to select one. It is standard practice to have the x-axis horizontal and the y-axis vertical and the equation "y= ax^2+ x+ c" is assuming that. You do have to decide where to put the origin of the coordinate system. I see two points where it would be reasonable to set the origin- at the initial point of the trajectory or and the end point. Choosing that initial point as the origin, x= 0 and y= 0 we must have y= 0= a(0^2)+ b(0)+c= c. So now we know the equation is y= ax^2+ x. The end point is 50 m down the hill which has a "slope of -3/4". The slope (not quite Calculus) is "y/x= -3/4" so y= -3x/4. Using the Pythagorean formula, x^2+ y^2= x^2+(3/4)^2x^2=(1+9/16)x^2= (25/16)x^2= 50^2 so x^2= (50^2)(16/25)= (2)(50)(16)=(4)(25)(16). Taking the square root of both sides, x= (2)(5)(4)= 40. y=-3x/4= -30. In coordinate system with origin at the starting point the end point has coordinates x= 40, y= -30. Putting those into y= ax^2+ x, -30=a(40)^2+ 40. -70= 1600a so a= -7/160.

With that freebie, condition for (a) is now met with y = (-7/160)*x^2 + x.
For (b), you merely have to complete the square (see Example 2, Section 2.6 of Swokowski's book) to determine the maximum height of the ball off the ground. No need for calculus but differentiation is somewhat easier and less work if you want to avoid all that "messy" completing the square stuff.

 

[Deleted member 4993]

Beer soaked hint follows.

With that freebie, condition for (a) is now met with y = (-7/160)*x^2 + x.
For (b), you merely have to complete the square (see Example 2, Section 2.6 of Swokowski's book) to determine the maximum height of the ball off the ground. No need for calculus but differentiation is somewhat easier and less work if you want to avoid all that "messy" completing the square stuff.

No messy work in locating the vertex of the parabola.

 

[allegansveritatem]

"g" is standard notation for the downward acceleration due to gravity and "sx" and "sy" are standard notation for x and y coordinates of position. However, this problem is giving you a different notation so you don't want to use those. In this problem the position of the ball is given by x and y and the acceleration is incorporated into "a".

In this problem, you are told that "using Calculus it can be shown that the path of the ball is given by y= ax^2+ x+ c for some constants a and c". So I take it you have not yet taken Calculus and are not expected to derive that. Since "real life" problems do not have a coordinate system attached you do need to select one. It is standard practice to have the x-axis horizontal and the y-axis vertical and the equation "y= ax^2+ x+ c" is assuming that. You do have to decide where to put the origin of the coordinate system. I see two points where it would be reasonable to set the origin- at the initial point of the trajectory or and the end point. Choosing that initial point as the origin, x= 0 and y= 0 we must have y= 0= a(0^2)+ b(0)+c= c. So now we know the equation is y= ax^2+ x. The end point is 50 m down the hill which has a "slope of -3/4". The slope (not quite Calculus) is "y/x= -3/4" so y= -3x/4. Using the Pythagorean formula, x^2+ y^2= x^2+(3/4)^2x^2=(1+9/16)x^2= (25/16)x^2= 50^2 so x^2= (50^2)(16/25)= (2)(50)(16)=(4)(25)(16). Taking the square root of both sides, x= (2)(5)(4)= 40. y=-3x/4= -30. In coordinate system with origin at the starting point the end point has coordinates x= 40, y= -30. Putting those into y= ax^2+ x, -30=a(40)^2+ 40. -70= 1600a so a= -7/160.

no, this is a problem from a precalculus text. I spent a long time yesterday re viewing parabolas and vertexes and their relationship. I then went back to the problem and calculated some more finding another point on the parabola. Still I was in a place where my wheels were still spinning in the sand...at least now it was sand they were spinning in and not deep mud. Later in the evening long after the math books had been stowed away it suddenly occurred to me that I could find one of the components of equation right off because one of the points I knew was 0,0. I could find c! This gave me great hope that with c out of the way I could build a system of equations with the other two components and Bob would soon be my uncle again! I have not yet done the work but today is the day. Therefore before I read the last posts in the thread I want to go back and see what I can do on my own. I can't actually say it is on my own however because Jomo's hint about the fact of the importance of there being a right triangle in the picture and Subotosh's suggestion for an approach to finding the equation by using the 0,0 were key pointers on the way. I am still not there but at least the car is moving. I will excuse myself here for foundering on the rocks of this easy problem by saying that for a long time I simply did not recognize the trajectory in the diagram as a parabola. It lacked the symmetry I associate with said figure. But by and by ( another eureka moment) I realized that it was indeed a parabola that had lost its left leg, so to speak, which had not been drawn in due to the nature of the situation being described in the problem. Once the parabolic nature of the figure was confirmed for me, the sky begin to open up. The rest, hopefully, will soon be history.

 

[allegansveritatem]

Beer soaked hint follows.

With that freebie, condition for (a) is now met with y = (-7/160)*x^2 + x.
For (b), you merely have to complete the square (see Example 2, Section 2.6 of Swokowski's book) to determine the maximum height of the ball off the ground. No need for calculus but differentiation is somewhat easier and less work if you want to avoid all that "messy" completing the square stuff.

no, this is a problem from a precalculus text. I spent a long time yesterday re viewing parabolas and vertexes and their relationship. I then went back to the problem and calculated some more finding another point on the parabola. Still I was in a place where my wheels were still spinning in the sand...at least now it was sand they were spinning in and not deep mud. Later in the evening long after the math books had been stowed away it suddenly occurred to me that I could find one of the components of equation right off because one of the points I knew was 0,0. I could find c! This gave me great hope that with c out of the way I could build a system of equations with the other two components and Bob would soon be my uncle again! I have not yet done the work but today is the day. Therefore before I read the last posts in the thread I want to go back and see what I can do on my own. I can't actually say it is on my own however because Jomo's hint about the fact of the importance of there being a right triangle in the picture and Subotosh's suggestion for an approach to finding the equation by using the 0,0 were key pointers on the way. I am still not there but at least the car is moving. I will excuse myself here for foundering on the rocks of this easy problem by saying that for a long time I simply did not recognize the trajectory in the diagram as a parabola. It lacked the symmetry I associate with said figure. But by and by ( another eureka moment) I realized that it was indeed a parabola that had lost its left leg, so to speak, which had not been drawn in due to the nature of the situation being described in the problem. Once the parabolic nature of the figure was confirmed for me, the sky begin to open up. The rest, hopefully, will soon be history.

 

[Deleted member 4993]

no, this is a problem from a precalculus text. I spent a long time yesterday re viewing parabolas and vertexes and their relationship. I then went back to the problem and calculated some more finding another point on the parabola. Still I was in a place where my wheels were still spinning in the sand...at least now it was sand they were spinning in and not deep mud. Later in the evening long after the math books had been stowed away it suddenly occurred to me that I could find one of the components of equation right off because one of the points I knew was 0,0. I could find c! This gave me great hope that with c out of the way I could build a system of equations with the other two components and Bob would soon be my uncle again! I have not yet done the work but today is the day. Therefore before I read the last posts in the thread I want to go back and see what I can do on my own. I can't actually say it is on my own however because Jomo's hint about the fact of the importance of there being a right triangle in the picture and Subotosh's suggestion for an approach to finding the equation by using the 0,0 were key pointers on the way. I am still not there but at least the car is moving. I will excuse myself here for foundering on the rocks of this easy problem by saying that for a long time I simply did not recognize the trajectory in the diagram as a parabola. It lacked the symmetry I associate with said figure. But by and by ( another eureka moment) I realized that it was indeed a parabola that had lost its left leg, so to speak, which had not been drawn in due to the nature of the situation being described in the problem. Once the parabolic nature of the figure was confirmed for me, the sky begin to open up. The rest, hopefully, will soon be history.

The equation of the trajectory is:

y = a * x^2 + x = x * (a*x + 1)

This is an equation of 2nd degree and it defines a parabola.

roots of this polynomial are x₁ = 0 and x₂ = -1/a

The vertex will be located at the mid-point of these roots. The y-coordinate of the vertex is the maximum height (off the x-axis) attained by the projectile.

 

[allegansveritatem]

Very quickly I upload what I did today. I will come back later and reply to posts I have not replied to. Let me preface by saying I know I missed it here but I couldn't find a way to construct a system with 0,0 as one of the points. Maybe someone can enlighten me on how to do that or why my whole approach is nuts. I used an approximate point, A pint I knew had to be pretty likely to be on or nearly on the actual trajectory of the ball.




I will be back a bit later. Thanks in advance to all who might comment.

 

[jonah2.0]

Tequila soaked hint follows.

As I said, condition for (a) has been met with y = (-7/160)*x^2 + x as posted by Sir HallsofIvy at post # 12. For (b), you merely have to complete the square for y = (-7/160)*x^2 + x (see Example 2, Section 2.6 of Swokowski's Precalculus book) to determine the maximum height of the ball off the ground.

Cheers.

 

[allegansveritatem]

Tequila soaked hint follows.

As I said, condition for (a) has been met with y = (-7/160)*x^2 + x as posted by Sir HallsofIvy at post # 12. For (b), you merely have to complete the square for y = (-7/160)*x^2 + x (see Example 2, Section 2.6 of Swokowski's Precalculus book) to determine the maximum height of the ball off the ground.

Cheers.

Here what I did for the vertex. No?:

I have looked at the Swokowski book section 2.6 long and hard in reviewing parabolas and vertices.