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**4 members have voted**

An airplane is directly 8 miles overhead from you, traveling at a speed of 600 mph. The plane maintains the same altitude and same direction. At that moment, you fire surface-to-air missile at it. The missile always travels at a speed of 2,000 mph and always points directly at the airplane.

How far away will the wreckage fall, assuming it will fall directly below the point of impact?

For full credit, I want to see an expression for the answer and solution.

Quote:WizardThis one is definitely worth of it's own thread. It pushes against the limits of my math ability.

An airplane is directly 8 miles overhead from you, traveling at a speed of 600 mph. At that moment, you fire surface-to-air missile at it. The missile always travels at a speed of 2,000 mph and always points directly at the airplane.

How far away will the wreckage fall, assuming it will fall directly below the point of impact?

For full credit, I want to see an expression for the answer and solution.

Here's my answer:

Dog Hand

Quote:DogHandHere's my answer:

Since you didn't specify the airplane's flight vector, I'll assume it is flying vertically away from you. Thus, the missile will strike the plane at a time of t = 8 miles/ (2000 - 600) mph = 1/175 hrs = 20.57... seconds, and the wreckage will fall right on top of you, killing you.

Dog Hand

Quote:WizardHow far away will the wreckage fall, assuming it will fall directly below the point of impact?

It isn't that hard to determine how far in terms of horizintal distance the airplane would travel.

8 miles = about 12.875 km

Assume the acceleration due to gravity is a constant 9.81 m/sec

^{2}

The time it takes to fall is t = sqrt(2 x 12,875 / 9.81) = about 51.23 seconds

The horizontal distance = 51.23 / 3600 hours x 600 miles/hour = about 8.54 miles.

It's a little bit above my normal math pay grade; I haven't worked with DEs in decades. Here's what I have so far:

Let x

_{t}and y

_{t}be the missile's location at time t in seconds; all distances are in miles

x

_{0}= y

_{0}= 0

At time t, the airplane's location is (600 t/3600, 8), or (t/6, 8)

The angle at time t, measured from the positive-x direction toward the positive-y direction, is Angle

_{t}= tan

^{-1}((8 - y

_{t}) / (t/6 - x

_{t})

The missile's velocity = 2000/3600 miles/second = 5/9

δy

_{t}/dt = 5/9 sin(Angle

_{t}) = 5/9 ((8 - y

_{t}) / sqrt((8 - y

_{t})^2 + (t/6 - x

_{t})^2)))

δx

_{t}/dt = 5/9 cos(Angle

_{t}) = 5/9 ((t/6 - x

_{t}) / sqrt((8 - y

_{t})^2 + (t/6 - x

_{t})^2)))

It would be easy to get a very good estimate of the answer in a spreadsheet. However, I'm looking for a mathematical solution.

Here is an integral you may need, which you could also easily look up.

1/sqrt(1+x

^{2}) dx = ln|x + sqrt(1+x

^{2})|

Quote:WizardCould it be I've stumped the math geniuses of the forum?

This appears to be the same question as the greyhound and the hare, which I remember working out in Calc 2 in college. Have not had time to sit down with it.

Quote:unJonThis appears to be the same question as the greyhound and the hare, which I remember working out in Calc 2 in college. Have not had time to sit down with it.

I will allow for extra time with this one. I spent a good part of the weekend on it, to be honest.

The spoiler will reveal the path of the missile and will give away the approximate answer. Don't click it if you don't want to know.

Predictable? One of the first pilots killed when Japanese fighter pilots dove into parachuting Americans was a pilot who died from a .38 calibre pistol wound!

Predictable? One of the first pilots killed when Japanese fighter pilots dove into parachuting Americans was a pilot who died from a .38 calibre pistol wound!