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CrystalMath
CrystalMath
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November 16th, 2020 at 11:06:56 PM permalink

1.9780219 miles
Purely numerical solution.
I don't even know where to start, mathematically. I'm sure it would have been no problem when I was in high school, but it makes my head hurt now.
I heart Crystal Math.
Wizard
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Wizard
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November 17th, 2020 at 5:32:07 AM permalink
Quote: CrystalMath


1.9780219 miles
Purely numerical solution.
I don't even know where to start, mathematically. I'm sure it would have been no problem when I was in high school, but it makes my head hurt now.



I've been wrong before, but I disagree. I worked out a numerical solution in Excel and you can see from the graph below, my answer at least 2.




It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
ThatDonGuy 
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November 17th, 2020 at 9:05:02 AM permalink
Quote: FleaStiff

i doubt any of the scatter pattern would be diretly below impact point, else airshows would be different.


It can't be, as the missile is always pointing toward the target. The only way the missile could be directly underneath the impact point is if the target was stationary and the missile was fired from directly underneath it.

I haven't gotten the "House lightbulb moment" for solving this one yet; I still have two partial derativates (where the horizontal and vertical velocity vectors are each in terms of the current horizontal and vertical location).

Quote: CrystalMath


1.9780219 miles
Purely numerical solution.
I don't even know where to start, mathematically. I'm sure it would have been no problem when I was in high school, but it makes my head hurt now.



...the Wizard's graph; using Excel, I get a contact point around 2.6375 miles down range.

Note that, assuming what's left of the plane maintains a forward velocity of 600 MPH, and assuming its vertical velocity is a constant 32 feet/sec^2, it will take about 51 seconds to hit the ground form an altitude of 8 miles, so its wreckage point is an additional 8.5 miles away.

Wizard
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Wizard
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November 17th, 2020 at 9:49:20 AM permalink
The next time I ask this, I won't ask about the distance to the wreckage, as I don't want to confuse the issue with the horizontal velocity of the wreckage.

Let the question be what is the horizontal distance from the missile launcher to the point on the ground directly below the point of impact eight miles high.

Quote: ThatDonGuy

I get a contact point around 2.6375 miles down range.



I agree. Still looking for a mathematical solution though.
It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
ThatDonGuy 
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November 17th, 2020 at 1:42:56 PM permalink
Quote: Wizard

I agree. Still looking for a mathematical solution though.


Still trying to figure one out.
gordonm888
gordonm888
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November 17th, 2020 at 2:42:40 PM permalink
Are we assuming the Earth is flat? Or do we need to account for curvature of the Earth?
So many better men, a few of them friends, were dead. And a thousand thousand slimy things lived on, and so did I.
Wizard
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Wizard
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November 18th, 2020 at 5:26:49 AM permalink
Quote: gordonm888

Are we assuming the Earth is flat? Or do we need to account for curvature of the Earth?



You may assume it's flat.

It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
ThatDonGuy 
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November 18th, 2020 at 7:24:35 AM permalink
Quote: ThatDonGuy

Still trying to figure one out.



...but I'm still trying to wrap my head around it. If there's an easier one, I'd like to see it.

However, the solution appears to be that the missile makes contact in 1440/91 seconds, so the airplane's horizontal distance is 1440/91 sec x 600/3600 miles/sec = 240/91 miles (about 2.63736),

Wizard
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Wizard
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November 18th, 2020 at 9:39:13 AM permalink
Quote: ThatDonGuy


...but I'm still trying to wrap my head around it. If there's an easier one, I'd like to see it.

However, the solution appears to be that the missile makes contact in 1440/91 seconds, so the airplane's horizontal distance is 1440/91 sec x 600/3600 miles/sec = 240/91 miles (about 2.63736),



I agree, it's 240/91. Still waiting to see a solution for full credit.
It's not whether you win or lose; it's whether or not you had a good bet.
Doc
Doc
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November 18th, 2020 at 12:12:42 PM permalink
I have to confess, I haven't even attempted to solve this problem. (I think I have become tired of my head frequently failing to do what it is supposed to be able to do, so maybe I avoid challenges!) When I first read the problem, my mind made a leap (assumption) that the missile path would be a parabola, but I don't really have much basis for that other than feeble "intuition."

I did like the looks of the Wizard's graph of the trajectory. While I was stationed at White Sands Missile Range fifty years ago, I watched quite a few launches, most of which were set to fly a programmed route, while a few were trying to intercept drones (testing the full system of missile, target-tracking system, missile-tracking system, and guidance). From ground level, I could not detect any visual difference between those paths and the Wizard's graph. About the only comment I can offer from that experience is that the assumption of a constant velocity missile would not represent the norm -- those things begin at rest, experience a period of very-high acceleration, and just maybe a period of constant velocity prior to intercept.

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