allegansveritatem
Full Member
- Joined
- Jan 10, 2018
- Messages
- 962
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.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!
The trajectory of the ball is: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
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.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]
In the picture (response #1) - the given plot of the trajectory started at point of intersection of the x and y axes (origin).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.
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.It has been forever since I tried to derive the formulas below. They look fine to be(!)
ax=0
vx = v0cos(theta)
sx = v0cos(theta)t + s0x
ay = -g
vy = -gt + v0sin(theta)
sy = -gt^2/2 + v0sin(theta)t + s0y
This will help you get your quadratic equation.
Note that to get the above formulas all I did was integrate ay = -g twice and ax= 0 twice.
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.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]
With that freebie, condition for (a) is now met with y = (-7/160)*x^2 + x."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 messy work in locating the vertex of the parabola.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."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.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.
The equation of the trajectory is: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.
Here what I did for the vertex. No?: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.