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Posers and Puzzles

Posers and Puzzles

  1. Standard member talzamir
    Art, not a Toil
    07 Nov '11 11:37
    An archer sees two monkeys 30 feet up a tree, 40 feet from the ground. His arrows fly are known with a starting velocity of v, and follow in flight a path shaped like a parabola, affected only by gravity - that is, their speed is sufficient that air friction does not play a significant role during the short flight.

    The first monkey just sits there on the branch.

    The second monkey is hanging from a branch. At the very instant when the arrow is released it lets go of the branch and starts falling down, with intent to land in the soft undergrowth under the tree, and then make a quick getaway.

    How will the archer hit the first monkey? How about the second monkey?
  2. Standard member talzamir
    Art, not a Toil
    07 Nov '11 13:27
    (err. 40 feet along the ground that is. seen from the side, (0,0) for the archer, (40,30) for the monkeys.)
  3. Standard member brang
    Lone star#
    07 Nov '11 17:20
    Originally posted by talzamir
    An archer sees two monkeys 30 feet up a tree, 40 feet from the ground. His arrows fly are known with a starting velocity of v, and follow in flight a path shaped like a parabola, affected only by gravity - that is, their speed is sufficient that air friction does not play a significant role during the short flight.

    The first monkey just sits there on the ...[text shortened]... make a quick getaway.

    How will the archer hit the first monkey? How about the second monkey?
    three pineapples
  4. Standard member brang
    Lone star#
    07 Nov '11 17:23
    throw an octopus at them
  5. 09 Nov '11 13:24 / 1 edit
    The second monkey is actually an easier case than the first. All the hunter need do is aim directly at the monkey at the moment before it falls from the branch.

    Gravity will affect the monkey and the arrow equally in terms of vertical acceleration - which is all that matters in this case. The initial height of the monkey and the horizantal velocity of the arrow will have no effect whatsoever on the final outcome.

    So long as the arrow is fired directly at the second monkey at the moment it begins falling, it will connect.


    As for the first monkey, I'm not sure. I'll have to leave it for someone with a bit more mathematical nouse.
  6. Standard member talzamir
    Art, not a Toil
    09 Nov '11 21:32
    You're right on that one. The other monkey is problematic. It gives me an equation of the form a sin x + b cos x + c = 0 where a, b and c are known and x to be solved.
  7. Subscriber Ponderable
    chemist
    10 Nov '11 13:06
    Originally posted by talzamir
    You're right on that one. The other monkey is problematic. It gives me an equation of the form a sin x + b cos x + c = 0 where a, b and c are known and x to be solved.
    let me throw a little bit of doubt into that. If the velocity of the arrow is not sufficient to hit the monkey before it hist the ground it won't hit. (Then there is of course no chance whatsoever to hit the other monkey at all
  8. Standard member talzamir
    Art, not a Toil
    11 Nov '11 07:24
    Hmm. It is entirely possible for a sin x + b cos x + c = 0 to have no solutions - just have a big enough c.

    Velocity of the arrow:
    * horizontal: v cos a
    * vertical: v sin a - gt

    Location of the arrow at time t:
    * horizontal: x = v cos a t
    * vertical: y = v sin a t - ½gt^2

    The time of impact, if there is such a thing, is when x = 40, so t = 40 / (v cos a). Then y = 30, which gives

    30 = v sin a * 40 / (v cos a) - ½g * (40 / (v cos a))^2

    which is a bit prettier as

    30 v^2 (cos a)^2 = 40v^2 * sin a cos a - 800 g

    15 v^2 - 15 v^2 cos 2x - 20v^2 sin 2x + 800 g = 0

    and there it is, a bit of rearrangement and it's a sin x + b cos x + c = 0.