not enough information

A

allegansveritatem

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Here is the problem (in two images because of page break):

12764
12765

It seems to me that they should tell the height reached also. The 45 degree angle factoid doesn't help in that this is a precalculus book and all about angles has not yet been introduced. So....what's a body to do with this problem? Here is what I came up with so far but I am far from convinced I got anything worth getting:

12766

There is one other thing that is a bit unusual about the information given: The form of the equation asked for is ax+x+c instead of ax+bx+c. I guess this has some important bearing on the question but I don't know exactly what at this point. One other thing: the reason I subtracted .642 from 175 is I thought it would make up for the fact that the cannon was 15 ft off the ground. But....I don't have a clear idea if this even makes sense.
 
y = ax^2 + x + c

The gift is that there is no 'b' attached to the linear term, x.

Using the point (0,15)

15 = a(0^2) + 0 + c = c, and this c = 15

y = ax^2 + x + 15

Using point (175,0)

0 = a(175^2) + 175 + 15 = 30625a + 190 ==> a = -38/6125 = -0.00620408163265306122448979591837

y = (-38/6125)x^2 + x + 15

You can now describe any point along the path. One useful point is the Vertex.

Since we already know (175,0), Solve this 0 = (-38/6125)x^2 + x + 15 (This is a little odd, since the other solution isn't actually on the path.)
OR
Since we already know (0,15), Solve this 15 = (-38/6125)x^2 + x + 15
To get a better handle on the vertex. It's in the middle.
 
You need to find a and b, you have twopoints and the equation. That really says it all.
 
tkhunny said:
y = ax^2 + x + c

The gift is that there is no 'b' attached to the linear term, x.

Using the point (0,15)

15 = a(0^2) + 0 + c = c, and this c = 15

y = ax^2 + x + 15

Using point (175,0)

0 = a(175^2) + 175 + 15 = 30625a + 190 ==> a = -38/6125 = -0.00620408163265306122448979591837

y = (-38/6125)x^2 + x + 15

You can now describe any point along the path. One useful point is the Vertex.

Since we already know (175,0), Solve this 0 = (-38/6125)x^2 + x + 15 (This is a little odd, since the other solution isn't actually on the path.)
OR
Since we already know (0,15), Solve this 15 = (-38/6125)x^2 + x + 15
To get a better handle on the vertex. It's in the middle.
Click to expand...
this is an interesting technique that I don't recall coming across in the text. Thanks. I will try to apply it tomorrow. I think part of my dilemma was the inclusion in the problem of the fact that the cannon was raked to a 45 degree angle. What would that have mattered when the problem was being given to people who had yet to learn what to do with such information? I guess I could have built some kind of right triangle....but I still wouldn't have had enough info to make anything of it.
 
Jomo said:
You need to find a and b, you have twopoints and the equation. That really says it all.
Click to expand...
I know that now...but as far as I can recall the only way I have learned to find an equation of a parabola involves knowing the vertex....anyway, I will investigate this method and post my results tomorrow. Thanks for posting
 
allegansveritatem said:
this is an interesting technique that I don't recall coming across in the text. Thanks. I will try to apply it tomorrow. I think part of my dilemma was the inclusion in the problem of the fact that the cannon was raked to a 45 degree angle. What would that have mattered when the problem was being given to people who had yet to learn what to do with such information? I guess I could have built some kind of right triangle....but I still wouldn't have had enough info to make anything of it.
Click to expand...
Occasionally, one is presented with a red herring.

The angle is used in a different sort of problem, such as "Can we get him to go farther if we use a different angle?"
 
tkhunny said:
y = ax^2 + x + c

The gift is that there is no 'b' attached to the linear term, x.

Using the point (0,15)

15 = a(0^2) + 0 + c = c, and this c = 15

y = ax^2 + x + 15

Using point (175,0)

0 = a(175^2) + 175 + 15 = 30625a + 190 ==> a = -38/6125 = -0.00620408163265306122448979591837

y = (-38/6125)x^2 + x + 15

You can now describe any point along the path. One useful point is the Vertex.

Since we already know (175,0), Solve this 0 = (-38/6125)x^2 + x + 15 (This is a little odd, since the other solution isn't actually on the path.)
OR
Since we already know (0,15), Solve this 15 = (-38/6125)x^2 + x + 15
To get a better handle on the vertex. It's in the middle.
Click to expand...
so, with the help of this post I went back at the thing and came up with this--which I think is not quite the same is OK anyway.:
12775

Right? or what?
 
allegansveritatem said:
so, with the help of this post I went back at the thing and came up with this--which I think is not quite the same is OK anyway.:
View attachment 12775

Right? or what?
Click to expand...
Right AND what! Very good.

I would have preferred a little more information about how you managed to find the vertex. Different piece of paper? Off the screen?
 
tkhunny said:
Right AND what! Very good.

I would have preferred a little more information about how you managed to find the vertex. Different piece of paper? Off the screen?
Click to expand...
What I did was to first find the equation as is put forth here and then I entered said equation into my excellent TI Nspire and used the graph trace tool to make sure that the parabola I got had the proper proportions, id est: y intercept of 15 and positive x intercept of 175. Then I traced along the parabola and found the vertex.
 
allegansveritatem said:
What I did was to first find the equation as is put forth here and then I entered said equation into my excellent TI Nspire and used the graph trace tool to make sure that the parabola I got had the proper proportions, id est: y intercept of 15 and positive x intercept of 175. Then I traced along the parabola and found the vertex.
Click to expand...
There you go. It's good to document methods. On to the next problem!!
 
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